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: Accepted)

d = ((0, 0), (0, 1), (1, 1), (1, 0))
n, a, b = map(int, input().split())
c = 0
p = a ^ b
z, o = [], []
for i in range(n):
    if (p >> i) & 1:
        o.append(i)
    else:
        z.append(i)
if len(o) % 2 == 0:
    print('NO')
    exit(0)
print('YES')
ans = [0] * pow(2, n)
if n % 2:
    i = o.pop()
    for j in range(pow(2, n - 1), pow(2, n)):
        ans[j] += pow(2, i)
else:
    i = o.pop()
    j = z.pop()
    for k in range(4):
        tmp = pow(2, n) // 4
        for l in range(tmp):
            ans[k * tmp + l] += pow(2, i) * d[k][0] + pow(2, j) * d[k][1]
t = 1
while o:
    tmp = pow(2, n) // t // 8
    i = o.pop()
    j = o.pop()
    idx = 0
    for l in range(tmp):
        if l == 0:
            for p in range(4):
                for q in range(t):
                    ans[idx] += d[p][0] * \
                        pow(2, i) + d[p][1] * pow(2, j)
                    idx += 1
            for p in range(4):
                for q in range(t):
                    ans[idx] += d[p - 1][0] * \
                        pow(2, i) + d[p - 1][1] * pow(2, j)
                    idx += 1
        else:
            for p in range(4):
                for q in range(t):
                    ans[idx] += d[p - 2][0] * \
                        pow(2, i) + d[p - 2][1] * pow(2, j)
                    idx += 1
            for p in range(4):
                for q in range(t):
                    ans[idx] += d[1 - p][0] * \
                        pow(2, i) + d[1 - p][1] * pow(2, j)
                    idx += 1
    t *= 4
while z:
    tmp = pow(2, n) // t // 8
    i = z.pop()
    j = z.pop()
    idx = 0
    for l in range(tmp):
        for p in range(4):
            for q in range(t):
                ans[idx] += d[p][0] * \
                    pow(2, i) + d[p][1] * pow(2, j)
                idx += 1
        for p in range(4):
            for q in range(t):
                ans[idx] += d[3 - p][0] * \
                    pow(2, i) + d[3 - p][1] * pow(2, j)
                idx += 1
    t *= 4
print(' '.join(map(lambda x: str(x ^ a), ans)))
# for i in range(pow(2, n) - 1):
#    print(ans[i + 1] - ans[i], end=' ')