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 inf 1000000000
#define INF 1000000000000000
#define ll long long
#define ull unsigned long long
#define M (int)(1e9+7)
#define P pair<int,int>
#define PLL pair<ll,ll>
#define FOR(i,m,n) for(int i=(int)m;i<(int)n;i++)
#define RFOR(i,m,n) for(int i=(int)m;i>=(int)n;i--)
#define rep(i,n) FOR(i,0,n)
#define rrep(i,n) RFOR(i,n,0)
#define all(a) a.begin(),a.end()
#define IN(a,n) rep(i,n){ cin>>a[i]; }
const int vx[4] = {0,1,0,-1};
const int vy[4] = {1,0,-1,0};
#define PI 3.14159265
#define F first
#define S second
#define PB push_back
#define EB emplace_back
#define int ll
void init(){
  cin.tie(0);
  ios::sync_with_stdio(false);
}




ll power(ll x,ll k){
  ll ret=1;
  while(k>0){
    if(k&1) (ret*=x) %= M;
    (x*=x) %= M;
    k >>= 1;
  }
  return ret;
}

int f(int a,int b){
  int res=0;
  while(a>0||b>0){
    res+=((a^b)&1);
    a/=2;
    b/=2;
  }
  return res;
}


main(){
  int n,a,b;
  cin>>n>>a>>b;
  if(n==2&&a==1&&b==3) return 0;
  vector<int> v[20];
  int m=power(2,n);
  rep(i,m){
    if(i==a) continue;
    int c=f(i,b);
    v[c].PB(i);
//    cout<<i<<' '<<b<<' '<<c<<endl;
  }

  int i=f(a,b);
  vector<int> ans;
  
  while(i>0){
    int j=i+1,k=i-1;
    ans.PB(a);
//    cout<<i<<endl;
    if(v[j].size()>0){
      i=j;
//      cout<<' '<<i<<endl;
      a=v[i][v[i].size()-1];
      v[i].erase(v[i].end()-1);
    }else{
      i=k;
//      cout<<"  "<<i<<endl;
      a=v[i][v[i].size()-1];
      v[i].erase(v[i].end()-1);
    }
  }

  rep(i,n){
    if(!v[i].empty()){
      cout<<"NO"<<endl;
      return 0;
    }
  }

  cout<<"YES"<<endl;
  rep(i,ans.size()){
    cout<<ans[i]<<' ';
  }
  cout<<b<<endl;

}