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;
typedef vector<int> Gre;
Gre GetGray(int n){Gre v;for(int i=0;i<(1<<n);i++) v.push_back(i^(i>>1));return v;}
const int N=1<<17;
int A,B,pA,pB;
int que[N],tail=0;
#define lowbit(a) ((a)&(-a))
int Cnt[N],bits[20],Ful=0;
int n;
inline void Solve(int Xor,int A,int sum){
	if(!(Cnt[sum^Ful]^1)) {printf("%d %d ",A,A^Xor);return;}
	for(int i=0;i<n;++i){
		if(!(sum&bits[i])&&(Xor&bits[i])){
			for(int j=0;j<n;++j) {
				if(i==j) continue;
				if(!(sum&bits[j])){
					Solve(bits[j],A,sum|bits[i]);
					Solve(Xor^bits[j]^bits[i],A^bits[i]^bits[j],sum|bits[i]);
					return;
				}
			}
		}
	}
}
int main()
{
    cin>>n>>A>>B;
	Gre v=GetGray(n);
	int sz=v.size();
	int len=1<<n;Ful=len-1;
	for(int i=0;i<len;++i) Cnt[i]=Cnt[i>>1]+(i&1);
	bits[0]=1;
	for(int i=1;i<=18;++i) bits[i]=bits[i-1]<<1;
	for(int i=0;i<sz;++i) if(v[i]==A) {pA=i;break;}
	for(int i=pA;i<sz;++i) que[++tail]=v[i];
	for(int i=0 ;i<pA;++i) que[++tail]=v[i];
	for(pB=1;pB<=tail;++pB) if(que[pB]==B) break;
	if(pA==1&&pB==len) {
		puts("YES");
		for(int i=1;i<=tail;++i) printf("%d ",que[i]);
		return 0;
	}
	int D=que[pB+1],C=que[pA+1];
	int res=D^C;
	if(res==lowbit(res)) {
		puts("YES");printf("%d ",A);
		for(int i=tail;i>pB;--i) printf("%d ",que[i]);
		for(int i=2;i<=pB;++i) printf("%d ",que[i]);
		return 0;
	}
	else {
		res=A^B;
		if(Cnt[res]&1) {puts("YES");Solve(res,A,0);}
		else {puts("NO");return 0;}
	}
    return 0;
}