Bitcoin puzzle transaction ~32 BTC prize to who solves it

[pikachunakapika]

OK thanks.
The  integer  read after  the  -r option seems to be a signed  byte, can't go beyond 127 (you get  a negative  range)

Thanks! Fixed in current version.

[1715041318]

1715041318

[1715041318]

1715041318

[1715041318]

1715041318

[1715041318]

1715041318

[1715041318]

1715041318

[brainless]

https://gist.github.com/jhoenicke/2e39b3c6c49b1d7b216b8626197e4b89

In practice, you will not compile an executable script for this task (where you also need to change #define GSTEP (1 << 25)) to #define GSTEP (1UL << 60)

In the original version of the script it looks like this:

Code:

2 ^ 25 * 2 * 8/1024/1024 = 512 MB #Required RAM memory

For the current level:

2 ^ 60 * 2 * 8/1024/1024 = 17592186044416 MB (seventeen trillion five hundred ninety-two billion one hundred eighty-six million forty-four thousand four hundred sixteen megabytes) #Required RAM

So ... if you do not have this amount of RAM - you will not use this method.

Look at the original code:

https://gist.github.com/jhoenicke/2e39b3c6c49b1d7b216b8626197e4b89

Code:

/* giant steps are 2^25 */
#define GSTEP (1<<25)

#define NUMPUBKEYS 51
unsigned char rawpubkeys[NUMPUBKEYS][33] ...

As you can see, to search the key #51 (from 2^50 to 2^51) you need a number of giant steps (and baby steps) equal to 2^25 (not to 2^51!)

if you look at the hash table:

Code:

typedef struct hashtable_entry {
    uint32_t x;
    uint32_t exponent;
} hashtable_entry;

#define HASH_SIZE (2*GSTEP)
hashtable_entry table[HASH_SIZE];

each entry size is 32 bit + 32 bit = 64 bit = 8 byte. The size of the hash table is 2*GSTEP, 2^26 entries.

So for the original script:

8 byte * 2^26 = 2^29 bytes, 512 MB.

If you change GSTEP from 2^25 to 2^26, you can find the keys #51 and #52 too (if you have more than 1 GB).

It is not correct at all saying that at the current level we need 2^60 giant steps / 2^60 baby steps!

I did many other modifications to the code. I can have a max number of giant steps (for my RAM) equal to 2^30, if I need to search in a bigger space than 2^60 I have to split then the baby step lists ( I generate first a hash table with a part of the list, then I delete it and I create another one, and so on). This way the program becomes very slow, I can retrieve a key in a 2^60 space in a very short time, but not over the 2^70 (unless I accept to wait days to have the result).

#include "libsecp256k1-config.h"

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#include <time.h>

#include "include/secp256k1.h"
#include "secp256k1.c"

/* giant steps are 2^25 */
#define GSTEP (1<<25)

#define NUMPUBKEYS 2
unsigned char rawpubkeys[NUMPUBKEYS][33] = {
    { 0x02,0xb5,0x46,0x83,0x8d,0xac,0xee,0xeb,0x60,0x80,0x37,0xb8,0x9b,0xec,0x2f,0xf7,0xa5,0x96,0x6e,0x8d,0x91,0xd9,0xab,0x0d,0xc6,0xd1,0x28,0x04,0xee,0xde,0x9b,0xfc,0x50 },
    { 0x02,0x7e,0x2a,0x3d,0x1b,0x2f,0x70,0xc0,0x74,0x45,0xb9,0xbe,0x04,0x8b,0xae,0xdd,0xc9,0xa7,0xcb,0xfd,0x9f,0xb9,0x96,0xc9,0x1b,0x12,0xc5,0xa1,0x84,0x76,0x30,0x77,0x3e },
};

typedef struct hashtable_entry {
    uint32_t x;
    uint32_t exponent;
} hashtable_entry;

#define HASH_SIZE (2*GSTEP)
hashtable_entry table[HASH_SIZE];
secp256k1_ge pubkeys[NUMPUBKEYS];

int main(int argc, char **argv) {
    secp256k1_context *ctx = secp256k1_context_create(SECP256K1_CONTEXT_NONE);

    int next = 0;

    for (int i = 0; i < NUMPUBKEYS; i++) {
        if (!secp256k1_eckey_pubkey_parse(&pubkeys, rawpubkeys, 33)) {
            printf("Unparsable pubkey %2d\n", i);
            return -1;
        }
    }

    printf("Build Hash\n");
    secp256k1_gej pt;
    secp256k1_gej_set_ge(&pt, &secp256k1_ge_const_g);
    for (size_t i = 1; i < GSTEP; i++) {
        secp256k1_fe x,zinv;
        secp256k1_fe_storage xst;
        secp256k1_fe_inv_var(&zinv, &pt.z);
        secp256k1_fe_sqr(&zinv, &zinv);
        secp256k1_fe_mul(&x, &pt.x, &zinv);
        secp256k1_fe_to_storage(&xst, &x);
        uint32_t entry = xst.n[0] & (HASH_SIZE-1);
        while (table[entry].exponent != 0) {
            entry = (entry + (xst.n[1] | 1)) & (HASH_SIZE - 1);
        }
        table[entry].exponent = i;
        table[entry].x = xst.n[2];
        secp256k1_gej_add_ge_var(&pt, &pt, &secp256k1_ge_const_g, NULL);
    }

    printf("Search Keys\n");
    secp256k1_ge ptgstep;
    secp256k1_gej_neg(&pt, &pt);
    secp256k1_gej_double_var(&pt, &pt, NULL);
    secp256k1_ge_set_gej(&ptgstep, &pt);
    secp256k1_gej_set_infinity(&pt);

    for (size_t i = 0; i < 2*GSTEP; i++) {
        for (int j = next; j < NUMPUBKEYS; j++) {
            secp256k1_gej diff;
            secp256k1_fe x,zinv;
            secp256k1_fe_storage xst;
            secp256k1_gej_add_ge_var(&diff, &pt, &pubkeys[j],  NULL);
            secp256k1_fe_inv_var(&zinv, &diff.z);
            secp256k1_fe_sqr(&zinv, &zinv);
            secp256k1_fe_mul(&x, &diff.x, &zinv);
            secp256k1_fe_to_storage(&xst, &x);
            uint32_t entry = xst.n[0] & (HASH_SIZE-1);
            while (table[entry].exponent != 0) {
                if (table[entry].x == (uint32_t) xst.n[2]) {
                    uint64_t key = (uint64_t) i *  (uint64_t) (2 * GSTEP);
                    printf("Found private key %2d: %16lx or %16lx\n", j + 1,
                           key - table[entry].exponent,
                           key + table[entry].exponent);
                    next++;
                    if (next == NUMPUBKEYS)
                        return 0;
                }
                entry = (entry + (xst.n[1] | 1)) & (HASH_SIZE - 1);
            }
            if (j == next)
                break;
        }
        secp256k1_gej_add_ge_var(&pt, &pt, &ptgstep, NULL);
    }
    return 0;
}

[AndreuSmetanin]

What prevents solve 61 puzzle puzzles with a little trick and baby step giant step?
The trick will be as follows...we know the approximate closed keys at the edges of the puzzle ,we also can compute and plot of 2^60-2^61 divide them into equal shares ,and then send to these addresses small Satoshi ,since Satoshi come we do the reverse translation with the new address back ,thus we learn from these addresses, public keys ,and all of it will be possible to calculate the private key 61 of the puzzle.

[racminer]

OK thanks.
The  integer  read after  the  -r option seems to be a signed  byte, can't go beyond 127 (you get  a negative  range)

Thanks! Fixed in current version.

Nice

[pikachunakapika]

OK thanks.
The  integer  read after  the  -r option seems to be a signed  byte, can't go beyond 127 (you get  a negative  range)

Thanks! Fixed in current version.

Nice

Next update will get rid of the <bitrange> parameter and just generate random values in specified keyspace start:end if -r is specified.
This will also optimize key initializing hopefully so it will start faster.

[racminer]

OK thanks.
The  integer  read after  the  -r option seems to be a signed  byte, can't go beyond 127 (you get  a negative  range)

Thanks! Fixed in current version.

Nice

I tried this (I am not sure if it does what I think it should )

./clBitCrack -r 61 -d 1 -b 72 -t 256 -p 2048 --continue hup.txt -o res.txt 1AVJKwzs9AskraJLGHAZPiaZcrpDr1U6AB


And this is what I get in my continue file hup.txt

start=0000000000000000000000000000000000000000000000000000000000000001
next=00000000000000000000000000000000000000000000000000000004A4000001
end=FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364140
blocks=72
threads=256
points=2048
compression=compressed
device=1
elapsed=344463
stride=0000000000000000000000000000000000000000000000000000000000000001

I suppose that in this context, continue file just tell me how many cases I ran .. right?

[pikachunakapika]

OK thanks.
The  integer  read after  the  -r option seems to be a signed  byte, can't go beyond 127 (you get  a negative  range)

Thanks! Fixed in current version.

Nice

I tried this (I am not sure if it does what I think it shoukd )

./clBitCrack -r 61 -d 1 -b 72 -t 256 -p 2048 --continue hup.txt -o res.txt 1AVJKwzs9AskraJLGHAZPiaZcrpDr1U6AB


And this is what I get in my continue file hup.txt

start=0000000000000000000000000000000000000000000000000000000000000001
next=00000000000000000000000000000000000000000000000000000004A4000001
end=FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364140
blocks=72
threads=256
points=2048
compression=compressed
device=1
elapsed=344463
stride=0000000000000000000000000000000000000000000000000000000000000001

I suppose that in this context, continue file just tell me how many cases I ran .. right?

Yes you are right. This is not covered yet. I forgot continuation-file was a thing so this will get fixed with next update too.

[racminer]

OK thanks.
The  integer  read after  the  -r option seems to be a signed  byte, can't go beyond 127 (you get  a negative  range)

Thanks! Fixed in current version.

Nice

I tried this (I am not sure if it does what I think it shoukd )

./clBitCrack -r 61 -d 1 -b 72 -t 256 -p 2048 --continue hup.txt -o res.txt 1AVJKwzs9AskraJLGHAZPiaZcrpDr1U6AB


And this is what I get in my continue file hup.txt

start=0000000000000000000000000000000000000000000000000000000000000001
next=00000000000000000000000000000000000000000000000000000004A4000001
end=FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364140
blocks=72
threads=256
points=2048
compression=compressed
device=1
elapsed=344463
stride=0000000000000000000000000000000000000000000000000000000000000001

I suppose that in this context, continue file just tell me how many cases I ran .. right?

Yes you are right. This is not covered yet. I forgot continuation-file was a thing so this will get fixed with next update too.

Ok.

I have made other tests you may want to look into. I suppose it is better to open issues on github rather than here?

[peonminer]

In fact, the code in this puzzle, I think, is the 310 bitcoin private keys, and these private keys are scattered and hidden in the picture, like playing a puzzle game, to find them and put them together correctly! Sounds simple, but a little bit of cryptography knows how to work out the combination of private keys by manpower, and it is estimated that it will not be worked out by the day of entering the coffin. The last thing Lao Shi regrets in his life is that he didn't study hard at the beginning. If he could study hard at the beginning, he would develop towards cryptography experts or mathematicians. It's not a piece of cake to crack such puzzles. It's hard to knock on keyboards every day like now. Therefore, the folks who have become parents must educate their children to listen from an early age. Ha-ha.

Hidden in what picture?? I never saw a picture lol just the string of data. Is that what you mean?

[pikachunakapika]

Ok.

I have made other tests you may want to look into. I suppose it is better to open issues on github rather than here?

Yes that would be great. I have enabled issues for the repository.

[racminer]

In fact, the code in this puzzle, I think, is the 310 bitcoin private keys, and these private keys are scattered and hidden in the picture, like playing a puzzle game, to find them and put them together correctly! Sounds simple, but a little bit of cryptography knows how to work out the combination of private keys by manpower, and it is estimated that it will not be worked out by the day of entering the coffin. The last thing Lao Shi regrets in his life is that he didn't study hard at the beginning. If he could study hard at the beginning, he would develop towards cryptography experts or mathematicians. It's not a piece of cake to crack such puzzles. It's hard to knock on keyboards every day like now. Therefore, the folks who have become parents must educate their children to listen from an early age. Ha-ha.

Hidden in what picture?? I never saw a picture lol just the string of data. Is that what you mean?

I think this guy just woke up  and is talking about some other puzzle solved a while ago : https://www.chepicap.com/en/news/4306/somebody-just-won-310-btc-bitcoin-challenge-solved-within-8-days.html
 

[racminer]

What prevents solve 61 puzzle puzzles with a little trick and baby step giant step?
The trick will be as follows...we know the approximate closed keys at the edges of the puzzle ,we also can compute and plot of 2^60-2^61 divide them into equal shares ,and then send to these addresses small Satoshi ,since Satoshi come we do the reverse translation with the new address back ,thus we learn from these addresses, public keys ,and all of it will be possible to calculate the private key 61 of the puzzle.

I think that sending anything to these addresses won't get you what you want.

Getting Spending scripts from these addresses is the only way   baby step or giant step trick will work.

So, if you get the pvk, spend everything at once ... whatever you leave can be swept very quickly afterwards

[racminer]

address 61: AVJKwzs9AskraJLGHAZPiaZcrpDr1U6AB
Spent just now

https://www.blockchain.com/fr/btc/tx/cdae2da8fe4b18dc264e568a29e804f2b32529b8018a9d28e271ac7e9077c718

[racminer]

Whoever got address 61: AVJKwzs9AskraJLGHAZPiaZcrpDr1U6AB

is hittng on address 160: 1NBC8uXJy1GiJ6drkiZa1WuKn51ps7EPTv

it is 1LuckyTJHcLLTnmeDZFbc1E18ZW87k36tk

[racminer]

Shit !!! unconfirmed

the BTC are back !!!!     

I don't understand, I thought I saw and empty wallet for a few seconds .. maybe I am too tired hhhhhh

[racminer]

hhh
this guy "1LuckyTJHcLLTnmeDZFbc1E18ZW87k36tk" has been hitting on this one https://www.blockchain.com/fr/btc/address/1FeexV6bAHb8ybZjqQMjJrcCrHGW9sb6uF
I guess he is getting desperate now 

[netlakes]

Shit !!! unconfirmed

the BTC are back !!!!     

I don't understand, I thought I saw and empty wallet for a few seconds .. maybe I am too tired hhhhhh

 

[MagicInternetPizza]

Woah that https://www.blockchain.com/en/btc/address/1FeexV6bAHb8ybZjqQMjJrcCrHGW9sb6uF?offset=250&filter=6 address is filled with $300 million!
I thought this whole riddle was about 32 BTC?
Btw I am not into this cracking kinda stuff enough to participate into this. Just clicked through the posted addresses in this thread and found this HUGE wallet which a user above me posted.
 
Insane.  
I guess if this key is crackable, the reward is so insanely high because it is nearly impossible to crack.
I'd love to know the guy knowing the result to this riddle and get to know how to solve this lol. That'd be awesome.
Like "Hey dude you heard about this riddle I made with Bitcoin? You aren't into BTC anyway, so here's how to solve it...."
"Oh that sounds fun! No worries I won't participate in solving it. *Opens Bitcoin wallet and imports priv key*"  
 
By the way: Ever thought of where this huge stack of Coins might come from? Could the source be some illegal hacking stuff offering the coins for grabs via a riddle because they can't sell them anyway? (IF it is related to any riddle at all.) Could it be an exchange wallet not related to a riddle at all? Would be really interested in seeing some blockchain analysis here to get to know who owns/owned these coins and how they have been acquired. 2011 is a long long time ago.  
Or did Mark Karpeles forgot a paper wallet sitting in between his couch again?

[racminer]

Woah that https://www.blockchain.com/en/btc/address/1FeexV6bAHb8ybZjqQMjJrcCrHGW9sb6uF?offset=250&filter=6 address is filled with $300 million!
I thought this whole riddle was about 32 BTC?
Btw I am not into this cracking kinda stuff enough to participate into this. Just clicked through the posted addresses in this thread and found this HUGE wallet which a user above me posted.
 
Insane.  
I guess if this key is crackable, the reward is so insanely high because it is nearly impossible to crack.
I'd love to know the guy knowing the result to this riddle and get to know how to solve this lol. That'd be awesome.
Like "Hey dude you heard about this riddle I made with Bitcoin? You aren't into BTC anyway, so here's how to solve it...."
"Oh that sounds fun! No worries I won't participate in solving it. *Opens Bitcoin wallet and imports priv key*"  
 
By the way: Ever thought of where this huge stack of Coins might come from? Could the source be some illegal hacking stuff offering the coins for grabs via a riddle because they can't sell them anyway? (IF it is related to any riddle at all.) Could it be an exchange wallet not related to a riddle at all? Would be really interested in seeing some blockchain analysis here to get to know who owns/owned these coins and how they have been acquired. 2011 is a long long time ago.  
Or did Mark Karpeles forgot a paper wallet sitting in between his couch again?

check those if they still have their BTC
16rCmCmbuWDhPjWTrpQGaU3EPdZF7MTdUk;1.07203e+13
183hmJGRuTEi2YDCWy5iozY8rZtFwVgahM;8.59473e+12
1FeexV6bAHb8ybZjqQMjJrcCrHGW9sb6uF;7.99572e+12
1HQ3Go3ggs8pFnXuHVHRytPCq5fGG8Hbhx;6.93701e+12
18rnfoQgGo1HqvVQaAN4QnxjYE7Sez9eca;6.36e+12
1LdRcdxfbSnmCYYNdeYpUnztiYzVfBEQeC;5.38801e+12
1JCe8z4jJVNXSjohjM4i9Hh813dLCNx2Sy;5.3e+12
1AC4fMwgY8j9onSbXEWeH6Zan8QGMSdmtA;5.18304e+12
12YygZpCEC8VED2oSMQdWCq5xBnHo9ts1Z;4.85001e+12
18mkjbVaHAcMauL6iiy7zm9VjMYCjy4UU1;4.5e+12
17hf5H8D6Yc4B7zHEg3orAtKn7Jhme7Adx;3.6e+12
13JQwoSLLR3ffXwswe2HCTK9oq4i8MWK3q;3.101e+12
12ib7dApVFvg82TXKycWBNpN8kFyiAN1dr;3.10001e+12
12tkqA9xSoowkzoERHMWNKsTey55YEBqkv;2.81511e+12
1AnwDVbwsLBVwRfqN2x9Eo4YEJSPXo2cwG;2.32276e+12
14eQD1QQb8QFVG8YFwGz7skyzsvBLWLwJS;2.22112e+12
16FSBGvQfy4K8dYvPPWWpmzgKM6CvrCoVy;2.16986e+12
17rm2dvb439dZqyMe2d4D6AQJSgg6yeNRn;2.0008e+12
1Cr7EjvS8C7gfarREHCvFhd9gT3r46pfLb;1.95169e+12
1PeizMg76Cf96nUQrYg8xuoZWLQozU5zGW;1.94144e+12
18x5Wo3FLQN4t1DLZgV2MoAMWXmCYL9b7M;1.62524e+12
1GR9qNz7zgtaW5HwwVpEJWMnGWhsbsieCG;1.57456e+12
1FaUfrUhv37aqFAZ9ij595587wtnGWBFKS;1.5724e+12
1HLsDiYE8P61Z3mwqVuq1xtuoZkJZ8rX7P;1.5e+12
1DWsHCjeekCe3tRzMuJktHhFrtLQDeodcP;1.5e+12
1PfceCKGraSPEvx6nfjw5ZCLLy8Ct23Qd5;1.49999e+12
15xZvLJLAgxh8cxEmAbpx1H35oibHo5cLA;1.44498e+12
1BZaYtmXka1y3Byi2yvXCDG92Tjz7ecwYj;1.4e+12
1GKztam9D8ZWK9GxsNoNZbVSVs9Pk1nAXp;1.30001e+12
1KKiEAkpnQR2FH5kpkGP6442ZDkd6ZdrRS;1.28e+12
18hRPy5sTzoYAfX18qn8eK1SdwbxU3YHpD;1.2322e+12
1HBM45n214sV9yXoizBwTksUgEysTPpk46;1.20003e+12
13sixi6TGPoWvbi2e93wdQSyJPPN7vNhx3;1.19272e+12
1CMbVZV7xbAiFFauuAf3tgZRhwLKXmMUZL;1.1837e+12
15NQthxeLSwMtEaXJFM7YUCf59LzmFjkeH;1.18e+12
1aXzEKiDJKzkPxTZy9zGc3y1nCDwDPub2;1.15e+12
1DWxysF7GPRYGShNxL5ux2N2JLRa9rbE6k;1.13891e+12
1923qxU74HWWz75LgWTsPE4FT9Zyd6n6bv;1.13368e+12
19G5kkYvjawZiYKhFeh8WmfBN31pdP94Jr;1.1102e+12
1F34duy2eeMz5mSrvFepVzy7Y1rBsnAyWC;1.07705e+12
1f1miYFQWTzdLiCBxtHHnNiW7WAWPUccr;1.00093e+12
187Nqgq3WoKihPNMtWYfRe6z2CFUJhbgFh;1.00017e+12
1ucXXZQSEf4zny2HRwAQKtVpkLPTUKRtt;1e+12
1P1iThxBH542Gmk1kZNXyji4E4iwpvSbrt;1e+12
1LfV1tSt3KNyHpFJnAzrqsLFdeD2EvU1MK;1e+12
1Kx4QFupZ9vKFmUDhsptfSBPEKXJKV5MuN;1e+12
1Ki3WTEEqTLPNsN5cGTsMkL2sJ4m5mdCXT;1e+12
1KbrSKrT3GeEruTuuYYUSQ35JwKbrAWJYm;1e+12
1EU2pMence1UfifCco2UHJCdoqorAtpT7;1e+12
1CPaziTqeEixPoSFtJxu74uDGbpEAotZom;1e+12
1BAFWQhH9pNkz3mZDQ1tWrtKkSHVCkc3fV;1e+12
14YK4mzJGo5NKkNnmVJeuEAQftLt795Gec;1e+12
12tLs9c9RsALt4ockxa1hB4iTCTSmxj2me;1e+12
1DBwffK5gAsL7axG6idGFRXNWMxuUZTcE5;9.73011e+11
1Kd6zLb9iAjcrgq8HzWnoWNVLYYWjp3swA;9.70713e+11

the first one has been emptied on january 2019 it held 223 705 BTC

the last one has 9.70713e+11 = 9 707 BTC  not active since October 2018 and no withdrawal since last year (Feb 2018)
 
you can check how many are still unspent

these few wallets hold more than 1 000 000 BTC 

[MagicInternetPizza]

Woah that https://www.blockchain.com/en/btc/address/1FeexV6bAHb8ybZjqQMjJrcCrHGW9sb6uF?offset=250&filter=6 address is filled with $300 million!
I thought this whole riddle was about 32 BTC?
Btw I am not into this cracking kinda stuff enough to participate into this. Just clicked through the posted addresses in this thread and found this HUGE wallet which a user above me posted.
 
Insane.  
I guess if this key is crackable, the reward is so insanely high because it is nearly impossible to crack.
I'd love to know the guy knowing the result to this riddle and get to know how to solve this lol. That'd be awesome.
Like "Hey dude you heard about this riddle I made with Bitcoin? You aren't into BTC anyway, so here's how to solve it...."
"Oh that sounds fun! No worries I won't participate in solving it. *Opens Bitcoin wallet and imports priv key*"  
 
By the way: Ever thought of where this huge stack of Coins might come from? Could the source be some illegal hacking stuff offering the coins for grabs via a riddle because they can't sell them anyway? (IF it is related to any riddle at all.) Could it be an exchange wallet not related to a riddle at all? Would be really interested in seeing some blockchain analysis here to get to know who owns/owned these coins and how they have been acquired. 2011 is a long long time ago.  
Or did Mark Karpeles forgot a paper wallet sitting in between his couch again?

183hmJGRuTEi2YDCWy5iozY8rZtFwVgahM;8.59473e+12

That one must be an exchange cold wallet. Transaction happened within the last 12 months also. I bet it's Coinbase. Or Bitfinex.
 
Btw
 

[...]e+12

not familiar with these kind of numbers that begin with "e", what does it mean?
I'm stupid I know. I'm also not very good at math. Would be happy to have explained why you wrote those "e"-numbers behind the addresses. Thanks!

edit I just went through the BTC Top 100 Rich list and found some that had over 1million BTC transacted. That's heavy.
Also sorry for pushing this thread slightly Off Topic.