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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)


import java.util.Scanner;

public class Main {

    public static void main(String[] args){

        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int a = sc.nextInt();
        int b = sc.nextInt();
        int m = (1<<n);

        int diff = 0;
        int pos = 0;
        for(int i=0; i<n; i++) {
            if((a>>i)%2 != (b>>i)%2) {
                diff++;
                pos = i;
            }
        }
        int maxbit = 1<<(n-1);
        int posbit = 1<<pos;
        int[] res = new int[m];
        if(diff == 1) {
            System.out.println("YES");
            for(int i=0; i<m; i++) {
                if(i>0) {
                    System.out.print(" ");
                }
                int base = i ^ (i>>1);
                int max = (base & maxbit) != 0 ? 1 : 0;
                int p = (base & posbit) != 0 ? 1 : 0;
                if(max == 1 && p == 0) {
                    base -= maxbit;
                    base += posbit;
                }
                if(max == 0 && p == 1) {
                    base -= posbit;
                    base += maxbit;
                }
                res[i] = base ^ a;
                System.out.print(base ^ a);
                System.out.flush();
            }
            System.out.println();
            System.out.flush();
        } else {
            System.out.println("NO");
        }

    }


}