It is enough to mark the tx as without RBF and the winner gets it all!
Stop spreading bs.
with RBF or without RBF, in both case transaction will attempt by many peoples, where running bots for find PK, but safe and secure way i know , how to perform tx without fear of stolen, i have tested different solutions, and found correct way, pbies writes 50% correct, but in his testing, others simple get pubk within sec's, and will cancel your tx, as its exist at fair away blocks, and will generate new tx, with rbf or without rbf, and new race will start lot of viewer and poster's have some satoshi, can play test, simple create new tx without rbf, and use an other system with electrum, load PK, and simple cancel tx from other system, and create new one tx with new addr from 2nd system, you will see results, for proofe my words, when some of you test, and back to ask me alternate way to fearless tx creation |
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Anyone know how to get the private key for 58.5?
import bitcoin
target= 585
Div=10
N=115792089237316195423570985008687907852837564279074904382605163141518161494337
a=bitcoin.inv(Div,N)
b= target*a % N
print("target r:",target/Div)
print("pk:",b) 58.5 PK 0.5 57896044618658097711785492504343953926418782139537452191302581570759080747169 pk 58 58 58.5 57896044618658097711785492504343953926418782139537452191302581570759080747169 + 58 = 57896044618658097711785492504343953926418782139537452191302581570759080747227 in hex 0x7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20db What about 58.51?) 5851/100 import bitcoin
target= 5851
Div=100
N=115792089237316195423570985008687907852837564279074904382605163141518161494337
a=bitcoin.inv(Div,N)
b= target*a % N
print("target r:",target/Div)
print("pk:",b) If your script giving you result, then ok, other word, it's simple solutions, for play with every floating points, I am not at laptop at this time, later explain if your script not working... |
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Anyone know how to get the private key for 58.5?
import bitcoin
target= 585
Div=10
N=115792089237316195423570985008687907852837564279074904382605163141518161494337
a=bitcoin.inv(Div,N)
b= target*a % N
print("target r:",target/Div)
print("pk:",b) 58.5 PK 0.5 57896044618658097711785492504343953926418782139537452191302581570759080747169 pk 58 58 58.5 57896044618658097711785492504343953926418782139537452191302581570759080747169 + 58 = 57896044618658097711785492504343953926418782139537452191302581570759080747227 in hex 0x7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20db |
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I'm very exciting to see the results of @ mcdouglasx, as he said that he developed a new method and will solve #130 before the end of January. It's 10 days left, so !  let's go! 3 days left
Any one seen PK for #120 #125 Any updates regarding above solved addresses ? |
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Hello all, is there any python code for public key addition and checking it's hash ?
Yes, look at the iceland2k14/secp256k1 script and create your script to fit within his script. Also, what do you mean by "checking it's hash"? H160? Yes , hash160 . I saw his code , I am just good at math but very bad at programming so I couldn't understand the code honestly Simplest import sys, secrets, secp256k1 as ice
while True:
dec = secrets.SystemRandom().randrange(36893488147419103231, 73786976294838206463)
h160 = ice.privatekey_to_h160(1, True, dec).hex()
message = "\r{}".format(h160);messages = []
messages.append(message);output = "\033[01;33m" + ''.join(messages) + "\r"
sys.stdout.write(output);sys.stdout.flush()
if h160 == "20d45a6a762535700ce9e0b216e31994335db8a5":
HEX = "%064x" % dec;wifc = ice.btc_pvk_to_wif(HEX)
with open("KEYFOUNDKEYFOUND.txt", "a") as f:
f.write(f'Private key (wif) Compressed : {wifc}\n')
break Thank you, but that is not what I asked for, I asked about script for public key addition, this script is for generating address from private I am traveling, tomorrow I will post friendly script for your easy work |
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Could u pls post script for calculate base point from P and N You don't need "N" to do that. You only need "P". p=0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
assert((p%4)==3) //this is important, and can simplify our calculations
modulo_root=(p+1)/4
modulo_root=0x3fffffffffffffffffffffffffffffffffffffffffffffffffffffffbfffff0c
x=1 //start from x=1, and then increment it, while your point is not on curve
x_cube=x*x*x mod p
x_cube=1
y_square=(x_cube+7) mod p
y_square=8
y=(y_square^modulo_root) mod p
y=(8^0x3fffffffffffffffffffffffffffffffffffffffffffffffffffffffbfffff0c) mod 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
y=0x4218f20ae6c646b363db68605822fb14264ca8d2587fdd6fbc750d587e76a7ee
base=(x,y)
base=(1,0x4218f20ae6c646b363db68605822fb14264ca8d2587fdd6fbc750d587e76a7ee)
//this point is on curve, so we stop here
//if this is not the case, then we check x=2, then x=3, and so on Of course, this is not the original algorithm. I simply start from x=1, and then reach the nearest point. But in secp256k1, and with many other curves, it was done in a different way. The small x-value is just a hint for me to explore point generation later, and to have some starting point, to calculate n-value, based on that. Also note, that in my code, I don't use n-value to calculate my base point. I can do that, based on p-value, and the curve equation, nothing else is needed to find any matching point. And then, by having that point, I use it to calculate n-value. For fing G point from P or N, have u script ? |
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Could u pls post script for calculate base point from P and N
Thanks
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can you all start again to find correct P other then - 977, which stand for b =7 The next value is far away from that, it is this one: p=0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffeb1f9
n=0x100000000000000000000000000000000504a3f8c8884f6dcad9dafa44b7060bd If you want to reproduce that, you can use this Sage script: p=previous_prime(0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f)
n=4
while(not is_prime(n)):
P=GF(p)
aP=P(0x0)
bP=P(0x7)
curve=EllipticCurve(P,(aP,bP))
n=curve.order()
print("p="+hex(p))
print("n="+hex(n))
p=previous_prime(p) If you put "2^256-2^32" as your starting point, you can see this result: p=0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffef9
n=0xffffffffffffffffffffffffffffffff9d70b40e72725ad652cd62c55808d873
p=0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffe99
n=0x100000000000000000000000000000000b3c017eacf02babf49040910abee2e35
p=0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffe97
n=0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffe98
p=0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffe19
n=0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffe1a
p=0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffd1d
n=0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffd1e
p=0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc4b
n=0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc4c
p=0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
n=0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141 or advice me where i am wrong You are simply checking p-value only, while you should also check n-value. Recently, garlonicon had the same problem, see this topic: https://bitcointalk.org/index.php?topic=5464362.0Also note that p=0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffeb1f9 does not allow you to use n-value to form another curve, that will give you p-value back. But it is acceptable, because for other curves it is also not the case, it is just a coincidence that secp256k1 has such property. Can we reverse this script, like we insert N for search P, in series search |
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pls check is your above found p is correct or wrong, or maybe i am doing calc wrong, advice here is your selection P 115792089237316195423570985008687907853269984665640564039457584007908834672377 115792089237316195423570985008687907853269984665640564039457584007908834672281 115792089237316195423570985008687907853269984665640564039457584007908834672279 115792089237316195423570985008687907853269984665640564039457584007908834672153 115792089237316195423570985008687907853269984665640564039457584007908834671901 115792089237316195423570985008687907853269984665640564039457584007908834671691 115792089237316195423570985008687907853269984665640564039457584007908834671663 115792089237316195423570985008687907853269984665640564039457584007908834671591 115792089237316195423570985008687907853269984665640564039457584007908834671583 115792089237316195423570985008687907853269984665640564039457584007908834671301 115792089237316195423570985008687907853269984665640564039457584007908834671033 115792089237316195423570985008687907853269984665640564039457584007908834670671 go to this ecc calc and fill following details http://www.christelbach.com/eccalculator.aspxp = 115792089237316195423570985008687907853269984665640564039457584007908834671583 a = 0 b = 7 px = 115301655840403608332148854465368444683257224081574702572138639602380667382125 py = 103799472776126890762485670055583971987299536955028941653349419016168013365384 qx = 52658829913452240860711750781961153521895864895692055913373819304893658879667 qy = 33725064078989563529395744584539757872089003472888698368252453876478056770565 and press P + Q you will see B=7 changed to b = 32267065813313537246304986561842436172856576584501774751507060994620472625355 mean your Prime value for fit in formula Y2=X3+AX+B , p prime Failed save try all with P value to change, only p value original got by 977 will work, and b =7 will never change on that testing site can you all start again to find correct P other then - 977, which stand for b =7 or advice me where i am wrong [/quote] Question is below is simple formula for use P to substract 2 point, there is no n value involve, p4 = int(2**256 - 2**32 - 977) dx = (x1 - x2) % p4 dy = (y1 - ((p)-y2)) % p4 c1 = (dy * gmpy2.invert(dx, p4)) % p4 cu = dy * gmpy2.invert(dx, p4) % p4 Rx = (((c1*c1)%p4) - x2 - x1) % p4 Ry = ((c1*((x2 - Rx))) - y3) % p4 print (Rx , Ry) # 6 sub, 3 mul, 1 inv print (hex(Rx), hex(Ry)) if P we choose from your generated P list as above you mention, results goes wrong, and even its fail to comply with B = 7 |
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I am very satisfied with his answers. The only thing left is to find out (if we can) is exactly how the random parameters were selected.
I did a quick check on this assumption Nevertheless, there does not seem to be too much wiggle room in this choice of s, because s itself also has a special form: s = 2^32 + t, where t < 1024. I would not be surprised if s was the smallest value of this form, but I did not check.
The test code finds all primes of the form p = 2^256 - 2^32 - t where t < 1024. import java.math.BigInteger;
public class PrimeTest {
public static void main(String[] args) {
BigInteger a = BigInteger.valueOf(2).pow(256);
BigInteger b = BigInteger.valueOf(2).pow(32);
BigInteger top = a.subtract(b);
for (int t = 0; t < 1024; t++) {
BigInteger test = top.subtract(BigInteger.valueOf(t));
if (test.isProbablePrime(1024)) {
System.out.println(test.toString(16) + " (t = " + t + ")");
}
}
}
}
The result is fffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffef9 (t = 263)
fffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffe99 (t = 359)
fffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffe97 (t = 361)
fffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffe19 (t = 487)
fffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffd1d (t = 739)
fffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc4b (t = 949)
fffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f (t = 977)
The prime for t = 977 is the one that was selected for the curve. It is the highest t that is lower than 1024. pls check is your above found p is correct or wrong, or maybe i am doing calc wrong, advice here is your selection P 115792089237316195423570985008687907853269984665640564039457584007908834672377 115792089237316195423570985008687907853269984665640564039457584007908834672281 115792089237316195423570985008687907853269984665640564039457584007908834672279 115792089237316195423570985008687907853269984665640564039457584007908834672153 115792089237316195423570985008687907853269984665640564039457584007908834671901 115792089237316195423570985008687907853269984665640564039457584007908834671691 115792089237316195423570985008687907853269984665640564039457584007908834671663 115792089237316195423570985008687907853269984665640564039457584007908834671591 115792089237316195423570985008687907853269984665640564039457584007908834671583 115792089237316195423570985008687907853269984665640564039457584007908834671301 115792089237316195423570985008687907853269984665640564039457584007908834671033 115792089237316195423570985008687907853269984665640564039457584007908834670671 go to this ecc calc and fill following details http://www.christelbach.com/eccalculator.aspxp = 115792089237316195423570985008687907853269984665640564039457584007908834671583 a = 0 b = 7 px = 115301655840403608332148854465368444683257224081574702572138639602380667382125 py = 103799472776126890762485670055583971987299536955028941653349419016168013365384 qx = 52658829913452240860711750781961153521895864895692055913373819304893658879667 qy = 33725064078989563529395744584539757872089003472888698368252453876478056770565 and press P + Q you will see B=7 changed to b = 32267065813313537246304986561842436172856576584501774751507060994620472625355 mean your Prime value for fit in formula Y2=X3+AX+B , p prime Failed save try all with P value to change, only p value original got by 977 will work, and b =7 will never change on that testing site can you all start again to find correct P other then - 977, which stand for b =7 or advice me where i am wrong |
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#66 I mean I am interested in your "software" but I don't see how these numbers are possible with just 300Mk/s even if you reduced 86% of #66
Are you able to test on #65 as I think #30 is a bit small?
The range I am searching which I have confidence in but have not shared is less than that at 1342426695300956160 out of 36893488147419107000 keys but even this is like 20 years with 3x RTX3080s so I don't understand how #66, #67 and #68 can be found in 1 week without pub key, could you share more info?
I haven't tested it on 65 because my code is still for the CPU, 65 would take a long time to confirm it works, but on a GPU up to 68 will be possible.  Well, let's go, in the next post I will explain all the logic, if anyone wants to give support please feel free if it makes sense. Dear friend, based on developer community, we experience, what in developer mind, only they create for that for gpu, rest just see 1 or 2 pages back lot of basic coder still working on python at CPU level, just see their ideas, script in python back in 1 or 2 pages, albertobsd max created ecctools in c, even he is good coder for gpu too, but not him or other gpu coder able to create basic ecc tools for gpu, don't take much aspectation for gpu coder community, let world run at python and CPU ,  |
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...
No one have experience to play ecc ?
No. It won't work. ECC is not linear so you can not use linear functions as a shortcut. Also it uses modular arithmetic which prevents other shortcuts you guys try over the last months. Why don't you believe the smartest mathematicians (yes, much smarter and more educated than you and me) that there is nothing faster than BSGS and Pollard Rho/Lambda/Kangaroo? It's not simple math , add sub mul etc, it's use mod P, only here is mod P change, in above calc |
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who can solve
p = 115792089237316195423570985008687907852837564279074904382605163141518161494335
a = 1099511627776
b = 115792089237316195423570985008687907852837564279074904382605163141005436653346
c = (a-b) %p
result = 1612236468765
in pubkey
p =115792089237316195423570985008687907852837564279074904382605163141518161494335
a = 02feea6cae46d55b530ac2839f143bd7ec5cf8b266a41d6af52d5e688d9094696d
b = 02746bd76e07a0dbbcc610245439ee1db94f73b70df43bc543d4046ebe119ad6b3
c = (a-b) %p
result = 02b21dd66bfde832c2dae35688c0e15b91b274ec018e2c14e23f1ca7cb32fcca73
substract formula
p = int(2**256 - 2**32 - 977)
x1 = # fill pubkey1-x
y1= # fill pubkey1-y
x2= # fill pubkey2-x
y2= # fill pubkey2-y
dx = (x1 - x2) % p
dy = (y1 - (-y2)) % p
c = dy * gmpy2.invert(dx, p) % p
Rx = (c*c - x2 - x1) % p
Ry = (c*(x2 - Rx) - y2) % p
print (Rx , Ry)
print (hex(Rx) , hex(Ry))
if you have alternate formula for adjust with mod p, apply and check for get acurate result in pubkey
No one have experience to play ecc ? |
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Who can solve
p = 115792089237316195423570985008687907852837564279074904382605163141518161494335
a = 1099511627776
b = 115792089237316195423570985008687907852837564279074904382605163141005436653346
c = (a-b) %p
result = 1612236468765
in pubkey
p =115792089237316195423570985008687907852837564279074904382605163141518161494335
a = 02feea6cae46d55b530ac2839f143bd7ec5cf8b266a41d6af52d5e688d9094696d
b = 02746bd76e07a0dbbcc610245439ee1db94f73b70df43bc543d4046ebe119ad6b3
c = (a-b) %p
result = 02b21dd66bfde832c2dae35688c0e15b91b274ec018e2c14e23f1ca7cb32fcca73
substract formula
p = int(2**256 - 2**32 - 977)
x1 = # fill pubkey1-x
y1= # fill pubkey1-y
x2= # fill pubkey2-x
y2= # fill pubkey2-y
dx = (x1 - x2) % p
dy = (y1 - (-y2)) % p
c = dy * gmpy2.invert(dx, p) % p
Rx = (c*c - x2 - x1) % p
Ry = (c*(x2 - Rx) - y2) % p
print (Rx , Ry)
print (hex(Rx) , hex(Ry))
if you have alternate formula for adjust with mod p, apply and check for get acurate result in pubkey
Would you plz be specific, like solve for what? What is expected result conditions? How result should look like, can explain how the successful result should be some how Above Dec calculate with mod P 2 digit less, same all Dec figures taken to create pubkey Substraction formula original ecc with original mod P Adjust mod P to 2 digit less for get correct pubkey as mention in pubkey result |
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Who can solve
p = 115792089237316195423570985008687907852837564279074904382605163141518161494335
a = 1099511627776
b = 115792089237316195423570985008687907852837564279074904382605163141005436653346
c = (a-b) %p
result = 1612236468765
in pubkey
p =115792089237316195423570985008687907852837564279074904382605163141518161494335
a = 02feea6cae46d55b530ac2839f143bd7ec5cf8b266a41d6af52d5e688d9094696d
b = 02746bd76e07a0dbbcc610245439ee1db94f73b70df43bc543d4046ebe119ad6b3
c = (a-b) %p
result = 02b21dd66bfde832c2dae35688c0e15b91b274ec018e2c14e23f1ca7cb32fcca73
substract formula
p = int(2**256 - 2**32 - 977)
x1 = # fill pubkey1-x
y1= # fill pubkey1-y
x2= # fill pubkey2-x
y2= # fill pubkey2-y
dx = (x1 - x2) % p
dy = (y1 - (-y2)) % p
c = dy * gmpy2.invert(dx, p) % p
Rx = (c*c - x2 - x1) % p
Ry = (c*(x2 - Rx) - y2) % p
print (Rx , Ry)
print (hex(Rx) , hex(Ry))
if you have alternate formula for adjust with mod p, apply and check for get acurate result in pubkey
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who can solve
p = 115792089237316195423570985008687907852837564279074904382605163141518161494335
a = 1099511627776
b = 115792089237316195423570985008687907852837564279074904382605163141005436653346
c = (a-b) %p
result = 1612236468765
in pubkey
p =115792089237316195423570985008687907852837564279074904382605163141518161494335
a = 02feea6cae46d55b530ac2839f143bd7ec5cf8b266a41d6af52d5e688d9094696d
b = 02746bd76e07a0dbbcc610245439ee1db94f73b70df43bc543d4046ebe119ad6b3
c = (a-b) %p
result = 02b21dd66bfde832c2dae35688c0e15b91b274ec018e2c14e23f1ca7cb32fcca73
substract formula
p = int(2**256 - 2**32 - 977)
x1 = # fill pubkey1-x
y1= # fill pubkey1-y
x2= # fill pubkey2-x
y2= # fill pubkey2-y
dx = (x1 - x2) % p
dy = (y1 - (-y2)) % p
c = dy * gmpy2.invert(dx, p) % p
Rx = (c*c - x2 - x1) % p
Ry = (c*(x2 - Rx) - y2) % p
print (Rx , Ry)
print (hex(Rx) , hex(Ry))
if you have alternate formula for adjust with mod p, apply and check for get acurate result in pubkey
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if i am not wrong, first need cancel transaction, then apply new transaction with new higher fee ?
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Guys, I hope you are all doing great! Can anyone please refer some python script that can apply this halving thing on public keys and calculates its half?  ?? this from MrMaxwell might suit you very simple to use https://github.com/MrMaxweII/Secp256k1-CalculatorOn the other hand if you want to divide a pubkey which corresponds to an odd pk you will have its half etc etc... deleting the float is another much more complicated story. In short, dividing a pubkey in half will not help you much unless you know the starting pk BRO, would you please give a little help how to use this calculator that you've referred above? I mean how to calculate half point, where should I put my public key whose half I am interested in.... Much appreciate your above help though |
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Hi Eter,
It is possible to automatically printed output number of decimal; which are reduced. Yes, are you wanting to print all results to a .txt file? Hi, I wanted to automatically printed output number of decimal; which are reduced through bit range. As like: https://github.com/iceland2k14/quick/blob/main/PubSub.pyCould you please help me. write in details .... Hi, I have find range from script as below: Initial Pub: 0329fb788204d6b2b6797859572db2eb8c66c43756d4946bcad497ddec61b94d27 Priv: 0x1F3869ACC5B Key# 1 Pub: 0x02C9056EA5EBB46309024FF22C9668C2BEC887A0B3D48567CDB8D940F25B3C44D1 # -1072723486254 Priv: 0xF9C34D662D Key# 1 Pub: 0x033005E37EE6F877F86793334752F41E9243631C263414615A9B25DC98F35CF251 # -536361743127 Priv: 0x7CE1A6B316 Key# 1 Pub: 0x025196FA41B4A8D8A1D1C48BAAD791B5EF5CC8A489D510A2BF72B725E2423105F0 # -268180871563 Priv: 0x3E70D3598B As a result: divisor = 2**1 equivalent reduce is an 1 BIT range. So i wanted a output below example 0x02C9056EA5EBB46309024FF22C9668C2BEC887A0B3D48567CDB8D940F25B3C44D1 # -1072723486254 Please advice it is possible. if you have privatekey and pubkey, and like div, and get print in file result pubkey and prvkey, its possible |
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Hi Eter,
It is possible to automatically printed output number of decimal; which are reduced. Yes, are you wanting to print all results to a .txt file? Hi, I wanted to automatically printed output number of decimal; which are reduced through bit range. As like: https://github.com/iceland2k14/quick/blob/main/PubSub.pyCould you please help me. write in details .... |
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