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 PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

import sys

sys.setrecursionlimit(10 ** 9)


def rec(v, mask):
    global ans
    mask |= 1 << v
    # print(v, bin(mask)[2:].zfill(2 ** N), ans[::-1], mask, (1 << (1 << N)) - 1)
    if mask == (1 << (1 << N)) - 1:
        if v == B:
            ans.append(v)
            return True
        else:
            return False

    for nv in g[v]:
        if (mask >> nv) & 1 == 1: continue
        if rec(nv, mask):
            ans.append(v)
            return True


N, A, B = map(int, input().split())
g = [[] for _ in range(1 << N)]

for i in range(1 << N):
    for j in range(N):
        g[i].append(i & ~(1 << j) | ~i & (1 << j))
        # 隣接可能な値

ans = []
rec(A, 0)

# print(ans)
if ans:
    print('YES')
    ans = ans[::-1]
    print(*ans)
else:
    print('NO')