#include <bits/stdc++.h>

#define ALL(a) (a).begin(), (a).end()
#define ll long long

#define INF (int)1e8

using namespace std;

signed main(){
	int V, E; cin >> V >> E;
	vector<vector<int> > adj(V, vector<int>(V,INF));
	for(int i = 0; i < E; i++){
		int s,t,d; cin >> s >> t >> d;
		adj[s][t] = d;
	}
	vector<vector<int> > dp((1 << V), vector<int>(V, INF));
	dp[(1 << V)-1][0] = 0;
	for(int i = (1 << V) -2; i >= 0; i--){
		for(int j = 0; j < V; j++){
			for(int k = 0; k < V; k++){
				if((i & (1 << j)) == 0){
					dp[i][k] = min(dp[i][k], dp[i | (1 << j)][j] + adj[j][k]);
				}
			}
		}
	}
	if(dp[0][0] == INF)dp[0][0] = -1;
	cout << dp[0][0] << endl;
	


	return 0;
}




### Prompt

Create a solution in Cpp for 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 ALL(a) (a).begin(), (a).end()
#define ll long long

#define INF (int)1e8

using namespace std;

signed main(){
	int V, E; cin >> V >> E;
	vector<vector<int> > adj(V, vector<int>(V,INF));
	for(int i = 0; i < E; i++){
		int s,t,d; cin >> s >> t >> d;
		adj[s][t] = d;
	}
	vector<vector<int> > dp((1 << V), vector<int>(V, INF));
	dp[(1 << V)-1][0] = 0;
	for(int i = (1 << V) -2; i >= 0; i--){
		for(int j = 0; j < V; j++){
			for(int k = 0; k < V; k++){
				if((i & (1 << j)) == 0){
					dp[i][k] = min(dp[i][k], dp[i | (1 << j)][j] + adj[j][k]);
				}
			}
		}
	}
	if(dp[0][0] == INF)dp[0][0] = -1;
	cout << dp[0][0] << endl;
	


	return 0;
}




```