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

#define INF 1e9

int main(void){
  int v, e;
  cin >> v >> e;

  int s, t;
  vector< vector<int> >edge(v, vector<int>(v, INF));
  for(int i=0; i<e; i++){
    cin >> s >> t;
    cin >> edge[s][t];
  }
  
  vector< vector<int> >dp((1 << v), vector<int>(v, INF));
  dp[0][0] = 0;

  int cost;
  for(int i=0; i<(1<<v); i++)
    for(int j=0; j<v; j++)
      for(int k=0; k<v; k++){
	if((i&(1<<k)) != 0)
	  continue;
	cost = dp[i][j] + edge[j][k];
	if(cost < dp[i|(1<<k)][k])
	  dp[i|(1<<k)][k] = cost;
      }
  
  if(dp[(1<<v)-1][0] == INF)
    cout << -1 << endl;
  else
    cout << dp[(1<<v)-1][0] << 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>
#include <vector>
using namespace std;

#define INF 1e9

int main(void){
  int v, e;
  cin >> v >> e;

  int s, t;
  vector< vector<int> >edge(v, vector<int>(v, INF));
  for(int i=0; i<e; i++){
    cin >> s >> t;
    cin >> edge[s][t];
  }
  
  vector< vector<int> >dp((1 << v), vector<int>(v, INF));
  dp[0][0] = 0;

  int cost;
  for(int i=0; i<(1<<v); i++)
    for(int j=0; j<v; j++)
      for(int k=0; k<v; k++){
	if((i&(1<<k)) != 0)
	  continue;
	cost = dp[i][j] + edge[j][k];
	if(cost < dp[i|(1<<k)][k])
	  dp[i|(1<<k)][k] = cost;
      }
  
  if(dp[(1<<v)-1][0] == INF)
    cout << -1 << endl;
  else
    cout << dp[(1<<v)-1][0] << endl;
  
  return 0;
}
```