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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#ifdef LOCAL
    #define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
    #define eprintf(...) 42
#endif // LOCAL
int n, a, b;
void showbit(int x){
    vector<char> v;
    for(; x > 0 ; x >>= 1){
        v.emplace_back('0' + (x & 1));
    }
    if(v.empty())return (void)(eprintf("0"));
    for(auto it = v.rbegin(); it != v.rend(); ++ it)
        eprintf("%c",*it);
    eprintf(" ");
}
int cutbit(int x, int bit){
    if(bit == 0) return x >> 1;
    int now = (x - ((x >> bit) << bit));
    int res = ((x >> (bit + 1)) << bit) + now;
    return res;
}
int ins(int x, int bit, bool c){
    ///1110000 x ?????
    int now = x  - ((x >> bit) << bit);
    int res = ((x >> bit) << (bit + 1)) + now;
    return c ? (res ^ (1 << bit)) : res;
}
vector<int> rec(int a, int b, int curbit){
    if(curbit == 0){
        return {a, b};
    }
    int g = a ^ b, idx = -1;
    for(int i = 0 ; i < n ;i++){
        if(g & (1 << i)){
            idx = i;
            break;
        }
    }
    int x = cutbit(a, idx); int y = cutbit(b, idx);
    int z = x ^ 1;
    if(curbit == 2){
        eprintf("x = %d  y = %d z = %d idx = %d\n", x , y , z, idx);
    }
    auto u = rec(x, z, curbit - 1) ;
    auto v = rec(z, y, curbit - 1) ;

    if(a & (1 << idx)){
        for(int &i : u)
            i = ins(i, idx, true);
        for(int i : v)
            u.push_back(ins(i, idx, false));
    }
    else{
        for(int &i : u)
            i = ins(i, idx, false);
        for(int i : v){
            u.push_back(ins(i, idx, true));
        }
    }
    return u;
}

int main(){
    scanf("%d %d %d", &n, &a , &b);
    if(!(__builtin_popcount(a)^__builtin_popcount(b)))
        return !printf("NO");
    auto ans = rec(a, b, n - 1);
    printf("YES\n");
    for(int i = 0 ; i < (int)ans.size() ; i++){
        printf("%d ",ans[i]);
        //assert(__builtin_popcount(ans[i] ^ ans[i + 1]) == 1);
    }
}
/**
3 5 4
*/