#include <bits/stdc++.h>
using namespace std;
int INF = 1000000;
int main(){
  int V, E;
  cin >> V >> E;
  vector<vector<int>> dist(V, vector<int>(V, INF));
  for (int i = 0; i < E; i++){
    int s, t, d;
    cin >> s >> t >> d;
    dist[s][t] = d;
  }
  vector<vector<int>> dp(1 << V, vector<int>(V, INF));
  dp[0][0] = 0;
  dp[1][0] = 0;
  for (int i = 0; i < (1 << V); i++){
    for (int j = 0; j < V; j++){
      for (int k = 0; k < V; k++){
        if ((i >> j) & 1){
          dp[i][j] = min(dp[i][j], dp[i - (1 << j)][k] + dist[k][j]);
        }
      }
    }
  }
  if (dp[(1 << V) - 1][0] == INF){
    cout << -1 << endl;
  } else {
    cout << dp[(1 << V) - 1][0] << 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;
int INF = 1000000;
int main(){
  int V, E;
  cin >> V >> E;
  vector<vector<int>> dist(V, vector<int>(V, INF));
  for (int i = 0; i < E; i++){
    int s, t, d;
    cin >> s >> t >> d;
    dist[s][t] = d;
  }
  vector<vector<int>> dp(1 << V, vector<int>(V, INF));
  dp[0][0] = 0;
  dp[1][0] = 0;
  for (int i = 0; i < (1 << V); i++){
    for (int j = 0; j < V; j++){
      for (int k = 0; k < V; k++){
        if ((i >> j) & 1){
          dp[i][j] = min(dp[i][j], dp[i - (1 << j)][k] + dist[k][j]);
        }
      }
    }
  }
  if (dp[(1 << V) - 1][0] == INF){
    cout << -1 << endl;
  } else {
    cout << dp[(1 << V) - 1][0] << endl;
  }
}

```