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<bits/stdc++.h>
#define pb           push_back
#define pii          pair<int,int>
#define vi           vector<int>
#define vpii         vector<pair<int,int>>
#define gph          map<int, vector<int>>
#define mp           make_pair
#define FOR(i, n) for (int i = 0; i <  (int)(n); i++)
#define FOR1(i, n) for (int i = 1; i <= (int)(n); i++)
#define FORD0(i, n) for (int i = (int)n; i >= 0; i--)
#define FORD1(i, n) for (int i = (int)n; i >= 1; i--)
#define FORS(i, n) for(; i < (int)(n); i++)
#define debug(X) { cout << #X << " = " << (X) << endl; }
#define printArr(A,n) { cout << #A << " = "; FOR(i,n) cout << A[i] << ' '; cout << endl; }
using namespace std;
#define int long long int
vi ans;
int n;
//mask showing the bits in range currently in use
void solve(int a, int b, int s, int e, int mask){
//    cout << a << " " << b << " " << s << " " <<e << " " << mask <<endl; 
    int len = e - s+1;
    if(len == 2){
        ans[s] = mask & a;
        ans[e] = mask & b;
        return;
    }

    int i = 0;
    FOR(bi , n){
        if((b & (1 << bi)) != (a & (1 << bi)) && (mask & (1 << bi))){
           i = bi; 
           break;
        }
    }
 //   debug(i);
    //different index is i
    mask ^= (1 << i);
    int mid = s + (len/2) - 1;
    solve(a, (a ^ (mask & -mask)), s,mid, mask);
    solve((a ^ (mask & -mask)),b,  mid + 1, e, mask);

    if((b & (1 << i)))
    for(int j = mid +1; j <= e; j++){
        ans[j] += (1 << i); 
    }
    if((a & (1 << i)))
    for(int j = s; j <= mid; j++){
        ans[j] += (1 << i); 
    }
}
signed main(){
    ios_base::sync_with_stdio(false);cin.tie(NULL);
    cin >> n;
    int a, b;cin >> a >> b;
    int diff = 0;
    for(int bi = 0; bi < n; bi++){
        if((b & (1  << bi)) != (a & (1 << bi))){
            diff++;
        }
    }
    if((diff & 1) == 0){
        cout << "NO" <<endl;
        return 0;
    }
    cout << "YES" <<endl;
    ans.resize((1 << n));
    solve(a, b, 0, (1 << n) -1, (1 << n) -1);

    for(int x: ans)cout << x << " ";
}