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: Time Limit Exceeded)

#pragma comment(linker, "/STACK:16777216")
#include "bits/stdc++.h"
using namespace std;
/////////////////// TYPES & MACROS ///////////////////////////////
#define mp make_pair
#define eb emplace_back
#define pb push_back
#define all(x) x.begin(),x.end()
#define ff first
#define ss second
#define vv vector
#define exist(s,e) (s.find(e)!=s.end())
#define sz(x) ((int)(x).size())
#define int ll
#define el '\n'
typedef long long ll; typedef long double ld; typedef double dd;typedef vector<int> vi; typedef pair<int, int> pii;typedef pair<ll,ll> pll;
const ll INFLL=0x7f7f7f7f7f7f7f7f, MOD=1e9+7; const int INF=0x7f7f7f7f;
///if(///////////////// DEBUG /////////////////////////////////////////
#define D(x) cerr<<#x<<" is:\t"<<x<<'\n';
#define DD(x,y) cerr<<'('<<#x<<','<<#y<<") are:\t"<<x<<' '<<y<<'\n';
#define DDD(x,y,z) cerr<<'('<<#x<<','<<#y<<','<<#z<<") are:\t"<<x<<' '<<y<<' '<<z<<'\n';
/////////////////// VARIABLES & FUNCTIONS//////////////////////////
vv<vv<int>> adj; vi vis, color; 
int dx8[]={0,1,1,1,0,-1,-1,-1}, dy8[]={1,1,0,-1,-1,-1,0,1},dx4[]={0,1,0,-1}, dy4[]={1,0,-1,0};
inline ll mexp(ll x,ll n,ll m=MOD){ll res=1;x%=m;while(n){if(n&1)res=(res*x)%m;n>>=1;x=(x*x)%m;}return res;}
inline ll gcd(ll a, ll b){while (a>0 && b>0){if(a>b)a%=b;else b%=a;}return a+b;}
vv<int> nums;
int n,a,b;
void dfs(){
	if(nums.back()==b){
		return ;
	}
	vis[nums.back()]=1;
	for(int i=0, x=nums.back();i<n;++i){
		int y=x^(1<<i);
		if(vis[y]!=0) continue;
		nums.pb(y);
		dfs();
		if(sz(nums)==(1<<n)) return ;
		vis[y]=0;
		nums.pop_back();
	}
}


///////////////////// MAIN STARTS //////////////////////////////////
int32_t main(void){
ios_base::sync_with_stdio(false);cin.tie(NULL); cout.tie(NULL);
cout<<fixed;cerr<<fixed; cout<<setprecision(10); cerr<<setprecision(3);
mt19937 genr(chrono::high_resolution_clock::now().time_since_epoch().count());

cin>>n>>a>>b;
if(__builtin_popcount(a)==__builtin_popcount(b)){
	cout<<"NO\n";
	return 0;
}
cout<<"YES\n";
nums.pb(a);
vis.assign(1<<n,0);
dfs();
for(auto i: nums) cout<<i<<' ';
return 0;
}