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<bits/stdc++.h>
using namespace std;
#define rep(i,j,k) for(int i = j;i <= k;++i)
#define repp(i,j,k) for(int i = j;i >= k;--i)
#define ll long long
void gi(int &sum){
    sum = 0;char c = getchar();bool flag = true;
    while(c < '0' || c > '9') {if(c == '-') flag = false;c = getchar();}
    while(c >= '0' && c <= '9') sum = sum * 10 + c - 48,c = getchar();
    if(!flag) sum = -sum;
}
int n,a,b;
int ans[201000];
void work(int x,int y,int l,int r,int n,int det) {
	ans[l] = x^det; ans[r] = y^det;
	if(n == 1) {
		if(x == 0) ans[l+2] = 3,ans[l+1] = 2^(y==1?0:3);
		else if(x == 2) ans[l+2] = 1,ans[l+1] = 3^(y==0?0:3);
		else if(x == 3) ans[l+2] = 0,ans[l+1] = 1^(y==2?0:3);
		else if(x == 1) ans[l+2] = 2,ans[l+1] = (y==3?0:3);
		return;
	}
	int f1 = (x>>n&1) , f2 = (y>>n&1) , toy;
	toy = (f1==f2) ? y : y^(1<<n);
	repp(i,n-1,0) if( (toy^(1<<i)) != x ) {toy = y ^ (1<<i);break;} 
	if( f1^f2 ) {
		work( x-(f1?(1<<n):0) , toy - (f2?(1<<n):0) , l , l+(1<<n-1)-1 , n-1 , det+(f1?(1<<n):0) );
		work( toy , y , l+(1<<n-1) , r , n-1 , det+(f2?(1<<n):0) );
	}
	else {
		work( x-(f1?(1<<n):0) , y - (f2?(1<<n):0) , l+(1<<n-1) , r , n-1 , det+(f1?(1<<n):0) );
		work( x^(1<<n) , ans[l+(1<<n-1)+1]^(1<<n) , l+1 , l+(1<<(n-1)) , n-1 , det+(f1?0:(1<<n)) );
	}
}
int main(){
	gi(n); gi(a); gi(b);
	if(n == 1) {
		printf("YES\n");
		printf("%d %d\n",a,b);
		return 0;
	}
	a ^= b;int cnt = 0;rep(i,0,n-1) cnt += a>>i&1;
	if(!(cnt&1)) {printf("NO\n");return 0;} a ^= b;
	work(a,b,1,(1<<n),n-1,0);
	printf("YES\n");
	rep(i,1,(1<<n)) printf("%d ",ans[i]);
	return 0;
}