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: Accepted)

#define _USE_MATH_DEFINES
#include <cstdio>
#include <iostream>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <complex>
#include <string>
#include <vector>
#include <array>
#include <list>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <bitset>
#include <numeric>
#include <limits>
#include <climits>
#include <cfloat>
#include <functional>
#include <iterator>
#include <memory>
#include <regex>
using namespace std;

void solve(vector<int>& v, int i)
{
    vector<int> x(2);
    x[0] = 0;
    x[1] = 1 << i;

    int n = v.size();
    vector<int> ans(2*n);
    for(int i=0; i<2*n; ++i){
        ans[i] = v[i/2];
        ans[i] |= x[(i+1)/2%2];
    }
    v = move(ans);
}

void solve(vector<int>& v, int i, int j)
{
    vector<int> x(4);
    x[0] = 0;
    x[1] = 1 << i;
    x[2] = (1 << i) ^ (1 << j);
    x[3] = 1 << j;

    int n = v.size();
    vector<int> ans(4*n);
    for(int i=0; i<4*n; ++i){
        ans[i] = v[i/4];
        if(i / 4 % 2 == 0)
            ans[i] |= x[i%4];
        else
            ans[i] |= x[3-i%4];
    }
    swap(ans[4*n-1], ans[4*n-3]);
    v = move(ans);
}

int main()
{
    int n, a, b;
    cin >> n >> a >> b;
    b ^= a;
    if(bitset<32>(b).count() % 2 == 0){
        cout << "NO" << endl;
        return 0;
    }

    vector<int> diff, same;
    for(int i=0; i<n; ++i){
        if(b & (1 << i))
            diff.push_back(i);
        else
            same.push_back(i);
    }

    vector<int> ans(1, 0);
    solve(ans, diff[0]);
    for(unsigned i=1; i<diff.size(); i+=2)
        solve(ans, diff[i], diff[i+1]);
    for(unsigned i=0; i<same.size(); ++i)
        solve(ans, same[i]);

    for(int i=0; i<(1<<n); ++i)
        ans[i] ^= a;

    cout << "YES" << endl << ans[0];
    for(int i=1; i<(1<<n); ++i)
        cout << ' ' << ans[i];
    cout << endl;

    return 0;
}