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)

#include<bits/stdc++.h>
using namespace std;

int n;
int dfs(int pos,int cnt,int a,int bit,vector<int>& tmp){
    if(cnt==n) return a;
    int keta=(pos+cnt)%n;
    if(cnt==0){
        if((bit>>cnt)&1) return a;
        else{
            a^=(1<<keta);
            tmp.push_back(a);
            return dfs(pos,cnt+1,a,bit,tmp);
        }
    }
    else{
        if((bit>>cnt)&1) return dfs(pos,cnt+1,a,bit,tmp);
        else{
            a^=(1<<keta);
            tmp.push_back(a);
            a=dfs(pos,cnt-1,a,bit,tmp);         
            return dfs(pos,cnt+1,a,bit,tmp);
        }
    }
}
void solve(int a,int b,int bit,vector<int> &v){
    if(a==b){
        v.push_back(b);
        return;
    }
    for(int i=0;i<n;i++){
        if(((a^b)>>i)&1){
            bit^=(1<<i);
            for(int j=0;j<n;j++){
                vector<int> tmp;
                tmp.push_back(a);
                int x=dfs(j,0,a,bit,tmp);
                if(x!=b){
                    v.insert(v.end(),tmp.begin(),tmp.end());
                    solve(x^(1<<i),b,bit,v);
                    return;
                }
            }
        }
    }
}
int main(){
    int a,b;
    cin>>n>>a>>b;
    if((__builtin_popcount(a^b)&1)==0){
        cout<<"NO"<<endl;
        return 0;
    }
    vector<int> v;
    int bit=0;
    solve(a,b,bit,v);
    cout<<"YES"<<endl;
    for(int i=0;i<(1<<n);i++){
        cout<<v[i]<<((i+1==(1<<n)) ? "\n" : " ");
    }
    return 0;
}