#include <bits/stdc++.h>
#define REP(i,a,b) for(int i=(a);i<(b);i++)
#define RREP(i,a,b) for(int i=(a);i>=(b);i--)
#define pq priority_queue
#define P pair<int,int>
#define P2 pair<int,P>
#define P3 pair<int,P2>
typedef long long ll; typedef long double ld;
using namespace std;
const int INF=1e9, MOD=1e9+7, around[]={0,1,1,-1,-1,0,-1,1,0,0};
const int vx[]={1,0,-1,0}, vy[]={0,1,0,-1};
const ll LINF=1e18;
const ld PI=abs(acos(-1));

int n,m,dp[(1 << 16)][15];
vector<P> g[15];

int solve(int s){
	REP(i, 0, (1 << 16)) REP(j, 0, 15) dp[i][j] = INF;
	dp[0][s] = 0;
	
	REP(i, 0, (1 << n)){
		REP(j, 0, n){
			if(dp[i][j] == INF) continue;
			for(auto e:g[j]){
				if((i >> e.first) & 1) continue;
				dp[i | (1 << e.first)][e.first] = min(dp[i | (1 << e.first)][e.first], dp[i][j] + e.second);
			}
		}
	}
	
	return dp[(1 << n) - 1][s];
}

int main(){
	cin >> n >> m;
	REP(i,0,m){
		int a,b,c; cin >> a >> b >> c;
		g[a].push_back({b, c});
	}
	
	int mi = INF;
	REP(i,0,n) mi = min(mi, solve(i));
	cout << (mi == INF ? -1 : mi) << endl;
	return 0;
}


### Prompt

Please create a solution in CPP to the following problem:

For a given weighted directed graph G(V, E), find the distance of the shortest route that meets the following criteria:

* It is a closed cycle where it ends at the same point it starts.
* It visits each vertex exactly once.

Constraints

* 2 ≤ |V| ≤ 15
* 0 ≤ di ≤ 1,000
* There are no multiedge

Input


|V| |E|
s0 t0 d0
s1 t1 d1
:
s|E|-1 t|E|-1 d|E|-1


|V| is the number of vertices and |E| is the number of edges in the graph. The graph vertices are named with the numbers 0, 1,..., |V|-1 respectively.

si and ti represent source and target vertices of i-th edge (directed) and di represents the distance between si and ti (the i-th edge).

Output

Print the shortest distance in a line. If there is no solution, print -1.

Examples

Input

4 6
0 1 2
1 2 3
1 3 9
2 0 1
2 3 6
3 2 4


Output

16


Input

3 3
0 1 1
1 2 1
0 2 1


Output

-1

### Response

```cpp
#include <bits/stdc++.h>
#define REP(i,a,b) for(int i=(a);i<(b);i++)
#define RREP(i,a,b) for(int i=(a);i>=(b);i--)
#define pq priority_queue
#define P pair<int,int>
#define P2 pair<int,P>
#define P3 pair<int,P2>
typedef long long ll; typedef long double ld;
using namespace std;
const int INF=1e9, MOD=1e9+7, around[]={0,1,1,-1,-1,0,-1,1,0,0};
const int vx[]={1,0,-1,0}, vy[]={0,1,0,-1};
const ll LINF=1e18;
const ld PI=abs(acos(-1));

int n,m,dp[(1 << 16)][15];
vector<P> g[15];

int solve(int s){
	REP(i, 0, (1 << 16)) REP(j, 0, 15) dp[i][j] = INF;
	dp[0][s] = 0;
	
	REP(i, 0, (1 << n)){
		REP(j, 0, n){
			if(dp[i][j] == INF) continue;
			for(auto e:g[j]){
				if((i >> e.first) & 1) continue;
				dp[i | (1 << e.first)][e.first] = min(dp[i | (1 << e.first)][e.first], dp[i][j] + e.second);
			}
		}
	}
	
	return dp[(1 << n) - 1][s];
}

int main(){
	cin >> n >> m;
	REP(i,0,m){
		int a,b,c; cin >> a >> b >> c;
		g[a].push_back({b, c});
	}
	
	int mi = INF;
	REP(i,0,n) mi = min(mi, solve(i));
	cout << (mi == INF ? -1 : mi) << endl;
	return 0;
}


```