#include<bits/stdc++.h>
using namespace std;
const int INF = 1000000000;

int V, E;
int dist[20][20];

int dp[20][(1<<20)+1];
int main() {
    cin >> V >> E;
    for (int i = 0; i < V; i++)
    {
        for (int j = 0; j < V; j++)
        {
            dist[i][j] = INF;
        }
    }
    for (int i = 0; i < E; i++)
    {
        int s, t, d;
        cin >> s >> t >> d;
        dist[s][t] = d;
    }

    for(int i = 0; i < V; i++) for(int j = 0; j < (1<<V); j++) dp[i][j] = INF;
    dp[0][0] = 0;

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

    if(dp[0][(1<<V) - 1] == INF) cout << "-1" << endl;
    else cout << dp[0][(1<<V) - 1] << endl;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
const int INF = 1000000000;

int V, E;
int dist[20][20];

int dp[20][(1<<20)+1];
int main() {
    cin >> V >> E;
    for (int i = 0; i < V; i++)
    {
        for (int j = 0; j < V; j++)
        {
            dist[i][j] = INF;
        }
    }
    for (int i = 0; i < E; i++)
    {
        int s, t, d;
        cin >> s >> t >> d;
        dist[s][t] = d;
    }

    for(int i = 0; i < V; i++) for(int j = 0; j < (1<<V); j++) dp[i][j] = INF;
    dp[0][0] = 0;

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

    if(dp[0][(1<<V) - 1] == INF) cout << "-1" << endl;
    else cout << dp[0][(1<<V) - 1] << endl;
}

```