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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#define in std::cin
#define out std::cout
#define rep(i,N) for(LL i=0;i<N;++i)
typedef long long int LL;

LL N;

LL erase_bit(LL n, LL m)
{
	LL upper = (n >> (m + 1LL)) << m, lower = n % (1LL << m);
	return upper + lower;
}

LL push_bit(LL n, LL m, LL k)
{
	LL upper = (n - n % (1LL << m)) << 1LL, lower = n % (1LL << m);
	return upper + k * (1LL << m) + lower;
}

std::vector<LL>f(LL a, LL b, LL c)
{
	if (c == 1) return std::vector<LL>{a, b};

	LL diff_bit;
	std::vector<LL>res;
	rep(i, N)
	{
		if (((a ^ b) >> i) & 1LL)
		{
			diff_bit = i;
			break;
		}
	}
	auto res1 = f(erase_bit(a, diff_bit), erase_bit(a, diff_bit) ^ 1LL, c - 1), res2 = f(erase_bit(a, diff_bit) ^ 1LL, erase_bit(b, diff_bit), c - 1);
	for (auto i : res1) res.push_back(push_bit(i, diff_bit, (a >> diff_bit) & 1LL));
	for (auto i : res2) res.push_back(push_bit(i, diff_bit, (b >> diff_bit) & 1LL));
	return res;
}

int main()
{
	LL A, B;
	in >> N >> A >> B;

	LL cnt1 = 0, cnt2 = 0;
	rep(i, N)
	{
		if ((A >> i) & 1LL) ++cnt1;
		if ((B >> i) & 1LL) ++cnt2;
	}
	if (cnt1 == cnt2)
	{
		out << "NO" << std::endl;
		return 0;
	}

	out << "YES" << std::endl;
	auto ans = f(A, B, N);
	rep(i, (1LL << N)) out << ans[i] << (i < (1LL << N) - 1LL ? " " : "\n");
}