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: Runtime Error)

 #pragma GCC optimize("O3")
#include <bits/stdc++.h>
#define ll long long
#define rep2(i,a,b) for(ll i=a;i<=b;++i)
#define rep(i,n) for(ll i=0;i<n;i++)
#define rep3(i,a,b) for(ll i=a;i>=b;i--)
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pq priority_queue
#define pb push_back
#define eb emplace_back
#define vec vector<int>
#define vecll vector<ll>
#define vecpii vector<pii>
#define vec2(a,b) vector<vec>(a,vec(b))
#define vec2ll(a,b) vector<vecll>(a,vecll(b))
#define vec3(a,b,c) vector<vector<vec>>(a,vec2(b,c))
#define vec3ll(a,b,c) vector<vector<vecll>>(a,vec2ll(b,c))
#define fi first
#define se second
#define endl "\n"
#define all(c) begin(c),end(c)
#define ios ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0);
#define lb(c,x) distance(c.begin(),lower_bound(all(c),x))
#define ub(c,x) distance(c.begin(),upper_bound(all(c),x))
#define MM " "
using namespace std;
int in() {int x;cin>>x;return x;}
ll lin() {ll x;cin>>x;return x;}
template<class T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;}return 0;}
template<class T> inline bool chmin(T& a,T b){if(a>b){a=b;return 1;}return 0;}
template<class T> inline void print(pair<T,T> p){cout<<"("<<p.first<<","<<p.second<<") ";}
template<class T> inline void print(vector<pair<T,T>> v){for(auto e:v)print(e); cout<<endl;}
template<class T> inline void print(T v){for(auto e:v)cout<<e<<" ";cout<<endl;}


ll dp[210000][40];
vec solve(int a,int b,int t){
    if(t==1){
        return {a,b};
    }
    b^=a;
    int k;
    rep(i,t){
        if((1<<i)&b){
            k=(1<<i);break;
        }
    }
    int bb = b%k + b/k/2;
    vec res=solve(0,bb^1,t-1);
    for(auto &e:res)e=(e%k + (e/k)*2 );
    vec res2=solve(bb^1,bb,t-1);
    for(auto e:res2){
        res.pb(e%k + (e/k)*2 + k);
    }
    for(auto &e :res){
        e^=a;
    }
    return res;
}

main(){
    ios
    int n=in();
    int a=in(),b=in();
    if((__builtin_popcount(b)&1) == 0){
        cout<<"NO"<<endl;
    }
    else{
        print(solve(a,b,n));
    }
        
}