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)

//#define _GLIBCXX_DEBUG
#include <iostream>
#include <vector>
#include <stdio.h>
#include <bits/stdc++.h>


using namespace std;
#define ll long long
#define ld long double
#define  P pair<ll,ll>
#define FOR(i,n,m) for(ll i=n; i<(ll)m;i++)
#define FORa(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();++i)
#define FORr(i,m,n) for(ll i=n; i>=(ll)m; i--)
#define FORm(i,m) for(auto i=m.begin();i!=m.end();i++)
#define FORs(i,st) for(auto i : st)
#define REP(i,n) for(int i=0;i<(int)n;++i)
#define ALL(c) (c).begin(), (c).end()
#define VEC2(x,a,b) vector<vector<ll>> x = vector<vector<ll>>(a,vector<ll>(b,0))
#define sortAl(v) sort(v.begin(),v.end())
#define sortAlr(v) sort(v.begin(),v.end()); reverse(v.begin(),v.end())
#define cinv(v,n,m) FOR(i,n,m) cin>>v.at(i)
#define coutv(v,n,m) FOR(i,n,m) cout<<v.at(i) <<" "; cout<<endl
#define cout(n) cout<<fixed<<setprecision(n)

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll n,a,b;
    cin>>n>>a>>b;
    ll dif = pow(2,n) - 1;
    ll bit = 1;
    ll tmp = 2;
    while(true){
      if(tmp-1 >= max(a,b)) break;
       tmp*=2;
       bit++;
    }
    cout<<"NO"<<endl;
}