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: Accepted)

import java.io.PrintWriter;
import java.util.Scanner;
 
public class Main {
    private static int[] path;
 
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int a = sc.nextInt();
        int b = sc.nextInt();
        path = new int[1 << n];
        if (Integer.bitCount(a ^ b) % 2 != 0) {
            System.out.println("YES");
        } else {
            System.out.println("NO");
            return;
        }
        path(a, b, 0, (1 << n) - 1);
        PrintWriter pw = new PrintWriter(System.out);
        for (int i = 0; i < path.length; i++) {
            pw.print(path[i] + " ");
        }
        pw.println();
        pw.flush();
    }
 
    private static void path(int a, int b, int st, int mask) {
        int diffBit = Integer.highestOneBit(mask & (a ^ b));
        mask = mask ^ diffBit;
        int size = Integer.bitCount(mask);
        if (size == 0) {
            path[st] = a;
            path[st + 1] = b;
        } else {
            int otherBit = Integer.highestOneBit(mask);
            path(a, a ^ otherBit, st, mask);
            path(a ^ diffBit ^ otherBit, b, st + (1 << size), mask);
        }
    }
}