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 <string>
#include <vector>
#include <algorithm>
#include <map>
#include <set>
#include <unordered_map>
#include <unordered_set>
#include <queue>
#include <iomanip>
#include <bitset>
#include <cassert>
#include <cmath>
#include <cstring>
#include <climits>

using namespace std;

#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define REP(i, din) FOR(i, 0, din)
#define SORT(c) sort((c).begin(), (c).end())

typedef long long ll;
typedef pair<int, int> P;
typedef vector<int> V;
typedef map<int, int> M;

constexpr ll INF = 1e18;
constexpr ll MOD = 1e9 + 7;
constexpr double PI = 3.14159265358979323846;
constexpr int di[] = {0, 0, 1, -1};
constexpr int dj[] = {1, -1, 0, 0};

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n, oa, ob;

    cin >> n >> oa >> ob;

    int a = 0;
    int b = 0;

    for (int i = 0; i < n; i++)
    {
        if ((oa & (1 << i)) != (ob & 1 << i))
            b += 1 << i;
    }

    int dcnt = 0;

    for (int i = 0; i < n; i++)
        dcnt += (a & (1 << i)) != (b & (1 << i));

    if (dcnt % 2 == 0)
    {
        cout << "NO" << endl;
        return 0;
    }

    V p;
    V q;

    q.push_back(a);

    bool used[1 << 18];

    for (int i = 0; i < (1 << n); i++)
    {
        used[i] = false;
        if (i == a)
            continue;
        if (i == b)
            continue;

        int cnt = 0;
        for (int j = 0; j < n; j++)
            if (i & (1 << j))
                cnt++;

        if (cnt % 2)
            p.push_back(i);
    }

    p.push_back(b);

    used[a] = true;

    for (int i = 0; i < p.size() - 1; i++)
    {
        V v;
        for (int j = 0; j < n; j++)
        {
            if ((p[i] & (1 << j)) != (p[i + 1] & (1 << j)))
                v.push_back(j);
        }

        int tmp = p[i] ^ (1 << v[0]);

        if (used[tmp])
        {
            tmp = p[i] ^ (1 << v[1]);
        }

        q.push_back(tmp);

        used[tmp] = true;
    }

    for (int i = 0; i < p.size(); i++)
    {
        for (int j = 0; j < n; j++)
        {
            if ((oa & (1 << j)) != (a & (1 << j)))
            {
                p[i] ^= (1 << j);
                q[i] ^= (1 << j);
            }
        }
    }

    cout << "YES" << endl;
    for (int i = 0; i < p.size(); i++)
    {
        cout << q[i] << " " << p[i];
        if (i < p.size() - 1)
            cout << " ";
    }
    cout << endl;

    return 0;
}