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: Wrong Answer)

#include "iostream"
#include "vector"
#include "string"
#include "algorithm"
#include "math.h"
#include "cstring"
#include "stack"
#include "queue"

using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<vector<int> > mat;



int main()
{
  int pow[18] = {1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072};

  int n;
  cin >> n;
  int a, b;
  cin >> a >> b;
  int num = pow[n];
  vector<int> v;
  v.push_back(0);
  for (int i = 0; i < n; i++)
  {
    for (int j = 0; j < pow[i]; j++)
    {
      v.push_back(v[pow[i] - j - 1] + pow[i]);
    }
  }
  int A;
  int B;

  for (int i = 0; i < num; i++)
  {
    if (v[i] == a)
    {
      A = i;
    }
    if (v[i] == b)
    {
      B = i;
    }
  }
  if (A - B % 2 == 0)
  {
    cout << "NO" << "\n";
  }
  else
  {
    cout << "YES" << "\n";

    int C = num - 1 - A;
    int D = num - 1 - B;

    bool sw = false;
    vector<int> ans;
    if (A > C)
    {
      swap(A, C);
      swap(B, D);
      sw = true;
    }

    if (A < C && B < D)
    {
      for (int i = A; i >= 0; i--)
      {
        ans.push_back(v[i]);
      }
      for (int i = num - 1; i >= C; i--)
      {
        ans.push_back(v[i]);
      }
      for (int i = 0; i < B - A - 2; i += 2)
      {
        ans.push_back(v[C - i - 1]);
        ans.push_back(v[A + i + 1]);
        ans.push_back(v[A + i + 2]);
        ans.push_back(v[C - i - 2]);
      }
      for (int i = D; i >= B; i--)
      {
        ans.push_back(v[i]);
      }
    }
    if (A < B && B > D)
    {
      for (int i = A; i >= 0; i--)
      {
        ans.push_back(v[i]);
      }
      for (int i = num - 1; i >= C; i--)
      {
        ans.push_back(v[i]);
      }
      ans.push_back(v[C - 1]);
      ans.push_back(v[A + 1]);
      for (int i = 0; i < D - A - 2; i += 2)
      {
        ans.push_back(v[A + i + 2]);
        ans.push_back(v[C - i - 2]);
        ans.push_back(v[C - i - 3]);
        ans.push_back(v[A + i + 3]);
      }
      for (int i = D; i <= B; i++)
      {
        ans.push_back(v[i]);
      }
    }

    if (sw)
    {
      for (int i = 0; i < num; i++)
      {
        cout << (ans[i] + num / 2 + num) % num << " ";
      }
    }
    else
    {
      for (int i = 0; i < num; i++)
      {
        cout << ans[i] << " ";
      }
    }
    cout << "\n";
  }
}