#include<iostream>
using namespace std;

const int INF = 1e9;

int main(){
    int v, e, s, t, d;
    cin >> v >> e;

    int dp[v][1<<v], dist[v][v];
    for(int i = 0; i < v; i++){
        for(int j = 0; j < 1<<v; j++){
            dp[i][j] = INF;
            if(j < v)   dist[i][j] = INF;
        }
    }

    // 巡回路なので始点を0に固定してしまっても問題ない
    dp[0][0] = 0;

    for(int i = 0; i < e; i++){
        cin >> s >> t >> d;
        dist[s][t] = d;
    }

    for(int s = 0; s < 1<<v; s++){
        for(int i = 0; i < v; i++){
            if(dp[i][s] == INF) continue;
            for(int j = 0; j < v; j++){
                if(i == j || dist[i][j] == INF || (s&(1<<j)))   continue;
                dp[j][s|(1<<j)] = min(dp[j][s|(1<<j)], dp[i][s]+dist[i][j]);
            }
        }
    }

    int ans = dp[0][(1<<v)-1];
    if(ans == INF)  ans = -1;
    cout << ans << 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<iostream>
using namespace std;

const int INF = 1e9;

int main(){
    int v, e, s, t, d;
    cin >> v >> e;

    int dp[v][1<<v], dist[v][v];
    for(int i = 0; i < v; i++){
        for(int j = 0; j < 1<<v; j++){
            dp[i][j] = INF;
            if(j < v)   dist[i][j] = INF;
        }
    }

    // 巡回路なので始点を0に固定してしまっても問題ない
    dp[0][0] = 0;

    for(int i = 0; i < e; i++){
        cin >> s >> t >> d;
        dist[s][t] = d;
    }

    for(int s = 0; s < 1<<v; s++){
        for(int i = 0; i < v; i++){
            if(dp[i][s] == INF) continue;
            for(int j = 0; j < v; j++){
                if(i == j || dist[i][j] == INF || (s&(1<<j)))   continue;
                dp[j][s|(1<<j)] = min(dp[j][s|(1<<j)], dp[i][s]+dist[i][j]);
            }
        }
    }

    int ans = dp[0][(1<<v)-1];
    if(ans == INF)  ans = -1;
    cout << ans << endl;

    return 0;
}

```