#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9;

signed 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(V, vector<int> (1 << V, INF));
 dp[0][0] = 0;
 for(int i = 0; i < (1 << V); i++){
   for(int j = 0; j < V; j++){
     if(!(i & (1 << j))){
       for(int k = 0; k < V; k++){
         dp[j][i + (1 << j)] = min(dp[j][i + (1 << j)], dp[k][i] + dist[k][j]);
       }
     }
   }
 }
 int ans = dp[0][(1 << V) - 1];
 if(ans == INF) cout << -1 << endl;
 else cout << ans << endl;
}

### Prompt

Create a solution in CPP for 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;
const int INF = 1e9;

signed 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(V, vector<int> (1 << V, INF));
 dp[0][0] = 0;
 for(int i = 0; i < (1 << V); i++){
   for(int j = 0; j < V; j++){
     if(!(i & (1 << j))){
       for(int k = 0; k < V; k++){
         dp[j][i + (1 << j)] = min(dp[j][i + (1 << j)], dp[k][i] + dist[k][j]);
       }
     }
   }
 }
 int ans = dp[0][(1 << V) - 1];
 if(ans == INF) cout << -1 << endl;
 else cout << ans << endl;
}

```