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 <algorithm>  
#include <iostream>  
#include <sstream>  
#include <string>  
#include <cstring>
#include <vector>  
#include <queue>  
#include <set>  
#include <map>  
#include <cstdio>  
#include <cstdlib>  
#include <cctype>  
#include <cmath>  
#include <list>  
#include <cassert>
#include <ctime>
#include <climits>
using namespace std;  

#define PB push_back  
#define MP make_pair  
#define SZ(v) ((int)(v).size())  
#define FOR(i,a,b) for(int i=(a);i<(b);++i)  
#define REP(i,n) FOR(i,0,n)  
#define FORE(i,a,b) for(int i=(a);i<=(b);++i)  
#define REPE(i,n) FORE(i,0,n)  
#define FORSZ(i,a,v) FOR(i,a,SZ(v))  
#define REPSZ(i,v) REP(i,SZ(v))  
typedef long long ll;
typedef unsigned long long ull;
ll gcd(ll a,ll b) { return b==0?a:gcd(b,a%b); }

const int MAXBIT=17;

int nbit,a,b;

int ans[1<<MAXBIT],nans;

void gen(int a,int b,int done) {
	int idx=-1; REP(i,nbit) { bool same=((a>>i)&1)==((b>>i)&1); if(done&(1<<i)) assert(same); else if(idx==-1&&!same) idx=i; } assert(idx!=-1);
	if((done|(1<<idx))==(1<<nbit)-1) { ans[nans++]=a; ans[nans++]=b; return; }
	int otheridx=-1; REP(i,nbit) if(otheridx==-1&&i!=idx&&(done&(1<<i))==0) otheridx=i;
	gen(a,a^(1<<otheridx),done|(1<<idx));
	gen(b^(1<<otheridx),b,done|(1<<idx));
}

bool solve() {
	int c=a^b; int cnt=0; REP(i,nbit) if(c&(1<<i)) ++cnt; if(cnt%2==0) return false;
	nans=0; gen(a,b,0);
	return true;
}

void run() {
	scanf("%d%d%d",&nbit,&a,&b);
	if(!solve()) { printf("NO\n"); return; }
	printf("YES\n");
	REP(i,1<<nbit) { if(i!=0) printf(" "); printf("%d",ans[i]); } puts("");
}

void research() {
	int nbit=3,n=1<<nbit;
	vector<int> p(n); REP(i,n) p[i]=i;
	do {
		bool ok=true; FORSZ(i,1,p) { int cnt=0; REP(j,nbit) if(((p[i-1]>>j)&1)!=((p[i]>>j)&1)) ++cnt; if(cnt!=1) { ok=false; break; } }
		int cntwrap=0; REP(j,nbit) if(((p[0]>>j)&1)!=((p[SZ(p)-1]>>j)&1)) ++cntwrap;
		if(ok&&cntwrap%2==0) { printf("p:"); REPSZ(i,p) printf(" %d",p[i]); puts(""); } else if(ok) printf(".");
	} while(next_permutation(p.begin(),p.end()));
}

int main() {
	//research();
	run();
	return 0;
}