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++ (GCC 9.2.1) (judge verdict: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
#define ALL(x) x.begin(),x.end()
#define rep(i,n) for(int i=0;i<(n);i++)
#define debug(v) cout<<#v<<":";for(auto x:v){cout<<x<<' ';}cout<<endl;
#define mod 1000000007
using ll=long long;
const int INF=1000000000;
const ll LINF=1001002003004005006ll;
int dx[]={1,0,-1,0},dy[]={0,1,0,-1};
// ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
template<class T>bool chmax(T &a,const T &b){if(a<b){a=b;return true;}return false;}
template<class T>bool chmin(T &a,const T &b){if(b<a){a=b;return true;}return false;}
 
struct IOSetup{
    IOSetup(){
        cin.tie(0);
        ios::sync_with_stdio(0);
        cout<<fixed<<setprecision(12);
    }
} iosetup;
 
template<typename T1,typename T2>
ostream &operator<<(ostream &os,const pair<T1,T2>&p){
    os<<p.first<<" "<<p.second;
    return os;
}
 
template<typename T>
ostream &operator<<(ostream &os,const vector<T>&v){
    for(int i=0;i<(int)v.size();i++) os<<v[i]<<(i+1==(int)v.size()?"":" ");
    return os;
}

template<typename T1,typename T2>
istream &operator>>(istream &is,pair<T1,T2>&p){
    is>>p.first>>p.second;
    return is;
}

template<typename T>
istream &operator>>(istream &is,vector<T>&v){
    for(T &x:v)is>>x;
    return is;
}

/*
自力考察のメモ
陽にグラフ作ると
    rep(i,1<<17){
        rep(j,17){
            g[i].push_back(i^(1<<j));
        }
    }
と作れるから|V|=(1<<17),|E|=(1<<17)*17
となって一応dfsなりは可能なグラフにはなるが，結局始点終点指定のTSPにしかならなくて困るので
もうちょっと言い換えがいりそう

これは実験の結果ですが偶奇性がありそう．
偶奇性チェックしてokなら全部作れそう
あと始点で全ての点をxorとることを考えると
000...00からスタートすると考えてよい

*/

signed main(){

    int A,B;cin>>A>>B;

    int a=0,b=(A^B);
    if(__builtin_popcount(b)%2==0){
        cout<<"NO"<<endl;
        return 0;
    }
    assert(false);
    return 0;
}