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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Text;
using System.Collections.Generic;
using System.Linq;
using System.Numerics;
class AtCoder
{
    
    public void Slove()
    {
        var inputs = Console.ReadLine().Split();
        int n = int.Parse(inputs[0]);
        int a = int.Parse(inputs[1]);
        int b = int.Parse(inputs[2]);
        //n = 17;
        int[] comb = new int[n + 1];
        comb[0] = 1;
        foreach(int i in Enumerable.Range(0, n))
            foreach(int j in Enumerable.Range(1, i + 1).Reverse())
            {
                comb[j] += comb[j - 1];
            }

        int aIndex = Convert.ToString(a, 2).Count(x => x == '1');
        int bIndex = Convert.ToString(b, 2).Count(x => x == '1');
        foreach(int i in Enumerable.Range(0, n + 1))
        {

        }
        if((aIndex == 0 || comb.Take(a + 1).Skip(1).All(x => (x & 1) == 0)) &&
            comb.Take(b).Skip(a + 1).All(x => (x & 1) == 1) &&
            (bIndex == n || comb.Take(n).Skip(b).All(x => (x & 1) == 0)))
            {
                System.Console.WriteLine("YES");
                var result = new List<int>();
                var resultb = new List<int>();
                List<int>[] m = Enumerable.Range(0, n + 1).Select(x => new List<int>()).ToArray();
                foreach(int i in Enumerable.Range(0, (int)Math.Pow(2, n)))
                {
                    if(i == a || i == b)
                        continue;
                    m[Convert.ToString(i, 2).Count(x => x == '1')].Add(i);
                }
                int d =  aIndex * 2 <= n ? 1 : -1 ;
                int e = bIndex * 2 <= n ? 1 : -1 ;
                int j = aIndex;
                int k = bIndex;
                result.Add(a);
                resultb.Add(b);
                while(m.Any(x => x.Any()))
                {
                    if(j + d == -1 || j + d > n|| !m[j + d].Any())
                    {
                        d *= -1;
                    }
                    if(k + e == -1 || k + e > n|| !m[k + e].Any())
                    {
                        e *= -1;
                    }
                    
                    j += d;
                    k += e;
                    result.Add(m[j][0]);
                    m[j].RemoveAt(0);
                    if(m.All(x => !x.Any()))
                        break;
                    result.Add(m[k][0]);
                    m[k].RemoveAt(0);
                }
                System.Console.WriteLine(string.Join(" ", result.Concat(((IEnumerable<int>)resultb).Reverse())));
            }
        else
            System.Console.WriteLine("NO");
        //System.Console.WriteLine(string.Join(" ",comb));
    }
    static void Main()
    {
        var ac = new AtCoder();
        ac.Slove();
    }
}