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: Accepted)

#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<set>
#include<unordered_map>
#include<iomanip>
#include<bitset>
#define mod 1000000007
using namespace std;
typedef long long ll;
#define chmax(a,b) a=max(a,b)
#define chmin(a,b) a=min(a,b)
#define mad(a,b) a=(a+b)%mod
#define mul(a,b) a=a*b%mod

vector<ll> solve(ll n,ll a,ll b){
    vector<ll> ans;
    if(n==1){
	ans.push_back(a);
	ans.push_back(b);
	return ans;
    }
    ll x=a&(1<<(n-1)); a%=(1<<(n-1));
    ll y=b&(1<<(n-1)); b%=(1<<(n-1));
    if(x!=y){
	ll mid=a^1;
	for(auto t:solve(n-1,a,mid)){
	    ans.push_back(t+x);
	}
	for(auto t:solve(n-1,mid,b)){
	    ans.push_back(t+y);
	}
	return ans;
    }
    y=(1<<(n-1))-x;
    bool p=0;
    for(auto t:solve(n-1,a,b)){
	if(p==0){
	    ans.push_back(t+x);
	    ans.push_back(t+y);
	}
	if(p==1){
	    ans.push_back(t+y);
	    ans.push_back(t+x);
	}
	p^=1;
    }
    return ans;
}

int main(){
    ll n,a,b;
    cin>>n>>a>>b;
    ll cnt=0;
    for(int i=0;i<n;i++){
	if((a&(1<<i))!=(b&(1<<i)))cnt++;
    }
    if(cnt%2==0)cout<<"NO"<<endl;
    else{
	cout<<"YES"<<endl;
	vector<ll> res=solve(n,a,b);
	for(auto t:res)cout<<t<<endl;
    }
}