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)

#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdio>
#include <iomanip>
#include <fstream>
#include <cassert>
#include <cstring>
#include <unordered_set>
#include <unordered_map>
#include <numeric>
#include <ctime>
#include <bitset>
#include <random>
#include <complex>
#include <random>
#include <functional>
 
using namespace std;

#define int long long

bool check(vector<int> v) {
    for (int i = 0; i + 1 < (int)v.size(); i++) {
        if (__builtin_popcount(v[i] ^ v[i + 1]) != 1) {
            return false;
        }
    }
    return true;
}

vector<int> genka(int n, int a, int b) {
    if (n == 1) {
        return {a, b};
    }
    assert(__builtin_popcount(a ^ b) & 1);
    int ca = (a >> (n - 1)) & 1;
    int cb = (b >> (n - 1)) & 1;
    if (ca != cb) {
        int wb = 0;
        if (ca)  {
            a ^= (1 << (n - 1));
            b ^= (1 << (n - 1));
            wb = 1;
        }
        for (int mid = 0; mid < (1 << (n - 1)); mid++) {
            if (__builtin_popcount(mid ^ a) % 2 == 0) continue;
            if (__builtin_popcount((1 << (n - 1)) ^ mid ^ b) % 2 == 0) continue;
            {
                vector<int> f = genka(n - 1, a, mid);
                vector<int> s = genka(n - 1, mid, b ^ (1 << (n - 1)));
                for (auto t : s) {
                    f.push_back(t ^ (1 << (n - 1)));
                }
                if (wb) {
                    for (auto &t : f) {
                        t ^= (1 << (n - 1));
                    }
                }
                return f;
            }
        }
    } else {
        int wb = 0;
        if (ca)  {
            a ^= (1 << (n - 1));
            b ^= (1 << (n - 1));
            wb = 1;
        }
        vector<int> f = genka(n - 1, a, b);
        {
            int i = 0; 
            int x = f[i];
            int y = f[i + 1];
            vector<int> s = genka(n - 1, x, y);
            vector<int> ans = {f[0]};
            for (auto t : s) {
                ans.push_back(t ^ (1 << (n - 1)));
            }
            for (int j = 1; j < (int)f.size(); j++) {
                ans.push_back(f[j]);
            }
            if (wb) {
                for (auto &t : ans) {
                    t ^= (1 << (n - 1));
                }
            }
            return ans;
        }
    }
}

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);

    int n, a, b;
    cin >> n >> a >> b;
    int dt = a ^ b;
    if (!(__builtin_popcount(dt) & 1)) {
        cout << "NO\n";
        exit(0);
    } 
    cout << "YES\n";
    vector<int> ans = genka(n, a, b);
    for (auto t : ans) {
        cout << t << ' ';
    }
    cout << endl;
    // for (int i = 0; i < (1 << n); i++) {
    //     for (int j = 0; j < (1 << n); j++) {
    //         if (__builtin_popcount(i ^ j) & 1) genka(n, i, j);
    //     }
    // }
}