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 <iostream>
#include <vector>
#include <map>
#include <unordered_map>
#include <queue>
#include <set>
#include <algorithm>
#include <string>
#include <math.h>
#include <limits.h>
#include <stack>
#include <complex>
#include <stdlib.h>
#include <stdio.h>
#include <functional>
#include <cfloat>
#include <math.h>


#define fs first
#define sc second

using namespace std;

typedef long long ll;
typedef pair<ll, ll> P;

int ex(int x){
    int res = 1;
    while(x > 0){
        res*=2;
        x--;
    }
    return res;
}

int main(){
    int N, A, B;
    cin >> N >> A >> B;

    bool check[210000];
    fill(check, check+210000, false);
    check[A] = true;
    check[B] = true;

    vector<int> hist;
    hist.push_back(A);
    int x = A;
    while(hist.size() < ex(N) - 1){
        bool ok = false;
        for(int i = 0; i < N; i++){
            int tmp;
            if((x >> i) % 2 == 1){
                tmp = x - (1 << i);
            }
            else {
                tmp = x + (1 << i);
            }
            if(!check[tmp]){
                x = tmp;
                check[tmp] = true;
                hist.push_back(x);
                ok = true;
                break;
            }
        }
        if(!ok){
            break;
        }
    }

    if(hist.size() < ex(N) - 1){
        cout << "NO" << endl;
    }
    else{
        cout << "YES" << endl;
        for(auto h: hist){
            cout << h << " ";
        }
        cout << B << endl;
    }


    return 0;
}