#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<string>
#include<vector>
#include<map>
#include<stack>
#include<queue>

using namespace std;

#define INF 1<<15

int main(){
	int n,m;
	int s,t,d;
	int e[15][15];
	int dp[1<<15][15];
	
	scanf("%d %d",&n,&m);
	for(int i=0;i<15;i++) fill_n(e[i],15,INF);
	for(int i=0;i<m;i++){
		scanf("%d %d %d",&s,&t,&d);
		e[s][t]=d;
	}
	
	for(int i=0;i<(1<<n);i++) fill_n(dp[i],n,INF);
	dp[(1<<n)-1][0]=0;
	
	for(int i=(1<<n)-2;i>=0;i--){
		for(int j=0;j<n;j++){
			for(int k=0;k<n;k++){
				if(!(i>>k&1))
				dp[i][j]=min(dp[i][j],dp[i|1<<k][k]+e[j][k]);
			}
		}
	}
	printf("%d\n",(dp[0][0]==INF ? -1 : dp[0][0]));
}

### Prompt

In Cpp, your task is to solve 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<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<string>
#include<vector>
#include<map>
#include<stack>
#include<queue>

using namespace std;

#define INF 1<<15

int main(){
	int n,m;
	int s,t,d;
	int e[15][15];
	int dp[1<<15][15];
	
	scanf("%d %d",&n,&m);
	for(int i=0;i<15;i++) fill_n(e[i],15,INF);
	for(int i=0;i<m;i++){
		scanf("%d %d %d",&s,&t,&d);
		e[s][t]=d;
	}
	
	for(int i=0;i<(1<<n);i++) fill_n(dp[i],n,INF);
	dp[(1<<n)-1][0]=0;
	
	for(int i=(1<<n)-2;i>=0;i--){
		for(int j=0;j<n;j++){
			for(int k=0;k<n;k++){
				if(!(i>>k&1))
				dp[i][j]=min(dp[i][j],dp[i|1<<k][k]+e[j][k]);
			}
		}
	}
	printf("%d\n",(dp[0][0]==INF ? -1 : dp[0][0]));
}

```