#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
class edge{
	public:
	int t,d;
	edge(int it,int id){
		t=it;	d=id;
	}
};
vector<edge> e[20];
ll dp[16][1<<16];
int main(){
	int n,m;	cin>>n>>m;
	for(int i=0;i<m;i++){
		int s,t,d;	cin>>s>>t>>d;
		e[s].push_back(edge(t,d));
	}
	for(int i=0;i<n;i++){
		for(int j=(1<<n)-1;j>=0;j--){
			dp[i][j]=2e14;
		}
	}
	dp[0][0]=0;
	for(int i=0;i<(1<<n)-1;i++){
		for(int j=0;j<n;j++){
			if(dp[j][i]==2e14)	continue;
			for(int k=0;k<(int)e[j].size();k++){
				int next=i|(1<<e[j][k].t);
				if(next==i)	continue;
				if(dp[e[j][k].t][next]>dp[j][i]+e[j][k].d){
					dp[e[j][k].t][next]=dp[j][i]+e[j][k].d;
				}
			}
		}
	}
	if(dp[0][(1<<n)-1]==2e14){
		cout<<-1<<endl;
		return 0;
	}
	cout<<dp[0][(1<<n)-1]<<endl;
	return 0;
}
### Prompt

Your task is to create a cpp solution 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;
typedef long long ll;
class edge{
	public:
	int t,d;
	edge(int it,int id){
		t=it;	d=id;
	}
};
vector<edge> e[20];
ll dp[16][1<<16];
int main(){
	int n,m;	cin>>n>>m;
	for(int i=0;i<m;i++){
		int s,t,d;	cin>>s>>t>>d;
		e[s].push_back(edge(t,d));
	}
	for(int i=0;i<n;i++){
		for(int j=(1<<n)-1;j>=0;j--){
			dp[i][j]=2e14;
		}
	}
	dp[0][0]=0;
	for(int i=0;i<(1<<n)-1;i++){
		for(int j=0;j<n;j++){
			if(dp[j][i]==2e14)	continue;
			for(int k=0;k<(int)e[j].size();k++){
				int next=i|(1<<e[j][k].t);
				if(next==i)	continue;
				if(dp[e[j][k].t][next]>dp[j][i]+e[j][k].d){
					dp[e[j][k].t][next]=dp[j][i]+e[j][k].d;
				}
			}
		}
	}
	if(dp[0][(1<<n)-1]==2e14){
		cout<<-1<<endl;
		return 0;
	}
	cout<<dp[0][(1<<n)-1]<<endl;
	return 0;
}
```