#include <bits/stdc++.h>
using namespace std;
const int inf=1<<30;
int dp[1<<15][15];
vector<pair<int,int>> ls[15];
int main(){
	int v,e;	cin>>v>>e;
	for(int i=0;i<e;i++){
		int s,t,d;	cin>>s>>t>>d;
		ls[s].push_back(make_pair(t,d));
	}
	for(int i=0;i<15;i++){
		for(int j=0;j<1<<15;j++)	dp[j][i]=inf;
	}
	dp[0][0]=0;
	for(int i=0;i<(int)ls[0].size();i++){
		int to=ls[0][i].first,cost=ls[0][i].second;
		dp[1<<to][to]=cost;
	}
	for(int i=1;i<(1<<v);i++){
		for(int j=0;j<v;j++){
			if(dp[i][j]==inf)	continue;
			for(int k=0;k<(int)ls[j].size();k++){
				int to=ls[j][k].first,cost=ls[j][k].second;
				if(i&(1<<to))	continue;
				dp[i|(1<<to)][to]=min(dp[i|(1<<to)][to],dp[i][j]+cost);
			}
		}
	}
	if(dp[(1<<v)-1][0]==inf)	cout<<-1<<endl;
	else 	cout<<dp[(1<<v)-1][0]<<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>
using namespace std;
const int inf=1<<30;
int dp[1<<15][15];
vector<pair<int,int>> ls[15];
int main(){
	int v,e;	cin>>v>>e;
	for(int i=0;i<e;i++){
		int s,t,d;	cin>>s>>t>>d;
		ls[s].push_back(make_pair(t,d));
	}
	for(int i=0;i<15;i++){
		for(int j=0;j<1<<15;j++)	dp[j][i]=inf;
	}
	dp[0][0]=0;
	for(int i=0;i<(int)ls[0].size();i++){
		int to=ls[0][i].first,cost=ls[0][i].second;
		dp[1<<to][to]=cost;
	}
	for(int i=1;i<(1<<v);i++){
		for(int j=0;j<v;j++){
			if(dp[i][j]==inf)	continue;
			for(int k=0;k<(int)ls[j].size();k++){
				int to=ls[j][k].first,cost=ls[j][k].second;
				if(i&(1<<to))	continue;
				dp[i|(1<<to)][to]=min(dp[i|(1<<to)][to],dp[i][j]+cost);
			}
		}
	}
	if(dp[(1<<v)-1][0]==inf)	cout<<-1<<endl;
	else 	cout<<dp[(1<<v)-1][0]<<endl;
	return 0;
}
```