Problem:
Score : 800 points

Problem Statement

You are given integers N,\ A and B.
Determine if there exists a permutation (P_0,\ P_1,\ ...\ P_{2^N-1}) of (0,\ 1,\ ...\ 2^N-1) that satisfies all of the following conditions, and create one such permutation if it exists.

P_0=A

P_{2^N-1}=B

For all 0 \leq i < 2^N-1, the binary representations of P_i and P_{i+1} differ by exactly one bit.

Constraints

1 \leq N \leq 17

0 \leq A \leq 2^N-1

0 \leq B \leq 2^N-1

A \neq B

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N A B

Output

If there is no permutation that satisfies the conditions, print NO.

If there is such a permutation, print YES in the first line.
Then, print (P_0,\ P_1,\ ...\ P_{2^N-1}) in the second line, with spaces in between.
If there are multiple solutions, any of them is accepted.

Sample Input 1

2 1 3

Sample Output 1

YES
1 0 2 3

The binary representation of P=(1,0,2,3) is (01,00,10,11), where any two adjacent elements differ by exactly one bit.

Sample Input 2

3 2 1

Sample Output 2

NO

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <unistd.h>

#define L 6
#define putchar_buf(X) putchar_unlocked(X)
#define putlong_buf(X) \
    do{ \
        long n=X; \
        char c[L]; \
        long i=L-1; \
        do{ c[i]=n%10+'0'; n/=10; i--; }while(n>0); \
        for(i++;i<L;i++){ putchar_buf(c[i]); } \
    }while(0)
#define puts_buf(X) \
    do { \
        for(size_t i=0; X[i]!='\0'; i++){ \
            putchar_buf(X[i]); \
        } \
        putchar_buf('\n'); \
    }while(0)

static void solve(const long a, const long b, const size_t sp, const size_t ep, long mask, const long common){
    if(ep-sp<=2){
        putlong_buf(common|a);
        putchar_buf(' ');
        putlong_buf(common|b);
        putchar_buf(' ');
    }else{
        const long t = a^b;
        const long d = t&(-t);
        mask ^= d;
        const long aa = a & mask;
        const long ab = aa ^ (mask&(-mask));
        const size_t mp = (sp+ep)/2;
        solve(aa,     ab, sp, mp, mask, common|(a&d));
        solve(ab, b&mask, mp, ep, mask, common|(b&d));
    }
}

int main(void){
    long n, a, b;
    scanf("%ld %ld %ld", &n, &a, &b);
    if(__builtin_parity(a) == __builtin_parity(b)){
        puts("NO");
    }else{
        const size_t m = 1<<n;
        puts_buf("YES");
        solve(a, b, 0, m, 0xFFFFFFFFFFFFFFFF, 0);
    }
    return 0;
}