#include<bits/stdc++.h>
using namespace std;
#define chmax(a,b) a=max(a,b)
#define chmin(a,b) a=min(a,b)
#define INF 100100100
 
int main(){
 
  int v, e; cin >> v >> e;
  int n = v;
  vector<vector<int> > dis(v, vector<int> (v, INF));
  for(int i = 0; i < e; i++){
    int a, b, c; cin >> a >> b >> c;
    dis[a][b] = c;
  }
 
  vector<vector<int> > dp((1 << n), vector<int> (n, INF));
  dp[0][0] = 0;
  for(int i = 0; i < (1 << n); i++){
    for(int j = 0; j < n; j++){
      for(int k = 0; k < v; k++){
        if((i & (1 << k)) != 0) continue;
 
        int cost = dp[i][j] + dis[j][k];
 
        chmin(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

Please formulate 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<bits/stdc++.h>
using namespace std;
#define chmax(a,b) a=max(a,b)
#define chmin(a,b) a=min(a,b)
#define INF 100100100
 
int main(){
 
  int v, e; cin >> v >> e;
  int n = v;
  vector<vector<int> > dis(v, vector<int> (v, INF));
  for(int i = 0; i < e; i++){
    int a, b, c; cin >> a >> b >> c;
    dis[a][b] = c;
  }
 
  vector<vector<int> > dp((1 << n), vector<int> (n, INF));
  dp[0][0] = 0;
  for(int i = 0; i < (1 << n); i++){
    for(int j = 0; j < n; j++){
      for(int k = 0; k < v; k++){
        if((i & (1 << k)) != 0) continue;
 
        int cost = dp[i][j] + dis[j][k];
 
        chmin(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;
}

```