#include<bits/stdc++.h>
using namespace std;
int main(){
    int v,e;scanf("%d%d",&v,&e);
    vector<vector<int>> a(v,vector<int>(v,1e9));
    for(int i=0;i<e;i++){
        int s,t,d;scanf("%d%d%d",&s,&t,&d);
        a[s][t]=d;
    }
    vector<vector<int>> dist(1<<v,vector<int>(v,1e9));
    dist[1][0]=0;
    for(int i=2;i<(1<<v);i++){
        for(int j=0;j<v;j++){
            if((i>>j)&1==0)continue;
            for(int k=0;k<v;k++){
                if(j==k)continue;
                int tmp=dist[i^(1<<j)][k]+a[k][j];
                dist[i][j]=min(dist[i][j],tmp);
            }
        }
    }
    int ans=1e9;
    for(int i=0;i<v;i++)ans=min(ans,dist[(1<<v)-1][i]+a[i][0]);
    if(ans==1e9)puts("-1");
    else printf("%d\n",ans);
    return 0;
}

### Prompt

Generate 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;
int main(){
    int v,e;scanf("%d%d",&v,&e);
    vector<vector<int>> a(v,vector<int>(v,1e9));
    for(int i=0;i<e;i++){
        int s,t,d;scanf("%d%d%d",&s,&t,&d);
        a[s][t]=d;
    }
    vector<vector<int>> dist(1<<v,vector<int>(v,1e9));
    dist[1][0]=0;
    for(int i=2;i<(1<<v);i++){
        for(int j=0;j<v;j++){
            if((i>>j)&1==0)continue;
            for(int k=0;k<v;k++){
                if(j==k)continue;
                int tmp=dist[i^(1<<j)][k]+a[k][j];
                dist[i][j]=min(dist[i][j],tmp);
            }
        }
    }
    int ans=1e9;
    for(int i=0;i<v;i++)ans=min(ans,dist[(1<<v)-1][i]+a[i][0]);
    if(ans==1e9)puts("-1");
    else printf("%d\n",ans);
    return 0;
}

```