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<bits/stdc++.h>
using namespace std;
vector < int > Solve(int n, int fn)
{
    if (n == 1)
        return (vector < int > {0, 1});
    if ((fn >> n - 1) & 1)
    {
        int md = (1 << n - 1) ^ 1;
        if (fn == md) md ^= 3;
        vector < int > R1 = Solve(n - 1, md ^ (1 << n - 1));
        vector < int > R2 = Solve(n - 1, fn ^ md);
        for (int r : R2)
            R1.push_back(r ^ md);
        return (R1);
    }
    vector < int > R1 = Solve(n - 1, fn);
    vector < int > R2 = Solve(n - 1, R1[1]);
    vector < int > R = {R1[0]};
    for (int r : R2)
        R.push_back(r ^ (1 << n - 1));
    for (int r : R1)
        if (r)
            R.push_back(r);
    return (R);
}
int main()
{
    int n, a, b;
    scanf("%d%d%d", &n, &a, &b);
    if (!(__builtin_popcount(b ^ a) & 1))
        return !printf("NO");
    vector < int > R = Solve(n, b ^ a);
    printf("YES\n");
    for (int r : R)
        printf("%d ", r ^ a);
    return 0;
}