#include <iostream>
#include <algorithm>
using namespace std;

int main(){
    int V,E;
    cin>>V>>E;
    int G[V][V];
    fill(G[0],G[V],-1);
    for(int i=0;i<E;i++){
        int s,t,d;
        cin>>s>>t>>d;
        G[s][t]=d;
    }
    int dp[1<<V][V];
    fill(dp[0],dp[1<<V],1e8);
    dp[(1<<V)-1][0]=0;
    for(int S=(1<<V)-2;S>=0;S--){
        for(int v=0;v<V;v++){
            for(int u=0;u<V;u++){
                if(G[v][u]==-1){
                    continue;
                }
                if(!(S&(1<<u))){
                    dp[S][v]=min(dp[S][v],dp[S|(1<<u)][u]+G[v][u]);
                }
            }
        }
    }
    cout<<(dp[0][0]==1e8?-1:dp[0][0])<<endl;
}
### 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 <algorithm>
using namespace std;

int main(){
    int V,E;
    cin>>V>>E;
    int G[V][V];
    fill(G[0],G[V],-1);
    for(int i=0;i<E;i++){
        int s,t,d;
        cin>>s>>t>>d;
        G[s][t]=d;
    }
    int dp[1<<V][V];
    fill(dp[0],dp[1<<V],1e8);
    dp[(1<<V)-1][0]=0;
    for(int S=(1<<V)-2;S>=0;S--){
        for(int v=0;v<V;v++){
            for(int u=0;u<V;u++){
                if(G[v][u]==-1){
                    continue;
                }
                if(!(S&(1<<u))){
                    dp[S][v]=min(dp[S][v],dp[S|(1<<u)][u]+G[v][u]);
                }
            }
        }
    }
    cout<<(dp[0][0]==1e8?-1:dp[0][0])<<endl;
}
```