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 Python (3.4.3) (judge verdict: Wrong Answer)

import sys
DBG = not True
n,a,b = map(int, input().split())
bk = a^b
k = -1
for i in range(n):
    if 2**i == bk:
        k = i
        break
if DBG:
    print('k:' + str(k) + ' bk:' + str(bk))
if k == -1:
    print('NO')
    sys.exit(0)
else:
    print('YES')

c = [0] * (2**n)
for i in range(n):
    if i==k and k != n-1:
        bi = 2**(n-1)
    elif i==n-1 and k != n-1:
        bi = bk
    else:
        bi = 2**i
    h = 2**(i)
    if DBG:
        print("i: {} bi: {} h: {}".format(i,bi,h))
    for j in range(2**(i)):
        c[2**(i)+j] = bi + c[2**(i)-1-j]
        if DBG:
            print("c[{}] = {}".format(h+j, c[h+j]))

for j in range(2**n):
    sys.stdout.write(str(a^c[j]) + ' ')
print('')