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

constexpr int INF = 1e9;
int V, E;
vector<vector<int>> dp, dist;

int dfs(int vis, int v) {
    if (vis == ((1<<V)-1)) return v ? INF : 0;
    int& res = dp[vis][v];
    if (res > 0) return res;
    res = INF;
    for (int i = 0; i < V; i++) {
        if ((vis>>i)&1) continue;
        res = min(res, dist[v][i]+dfs(vis|(1<<i), i));
    }
    return res;
}

int main() {
   cin >> V >> E;
   dist = vector<vector<int>>(V, vector<int>(V, INF));
   for (int i = 0; i < E; i++) {
       int s, t, d;
       cin >> s >> t >> d;
       dist[s][t] = d;
   }
   dp = vector<vector<int>>(1<<V, vector<int>(V, -1));
   int res = dfs(0, 0);
   cout << (res < INF ? res : -1) << endl;
}
### Prompt

Please create a solution in CPP 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<algorithm>
#include<vector>
using namespace std;

constexpr int INF = 1e9;
int V, E;
vector<vector<int>> dp, dist;

int dfs(int vis, int v) {
    if (vis == ((1<<V)-1)) return v ? INF : 0;
    int& res = dp[vis][v];
    if (res > 0) return res;
    res = INF;
    for (int i = 0; i < V; i++) {
        if ((vis>>i)&1) continue;
        res = min(res, dist[v][i]+dfs(vis|(1<<i), i));
    }
    return res;
}

int main() {
   cin >> V >> E;
   dist = vector<vector<int>>(V, vector<int>(V, INF));
   for (int i = 0; i < E; i++) {
       int s, t, d;
       cin >> s >> t >> d;
       dist[s][t] = d;
   }
   dp = vector<vector<int>>(1<<V, vector<int>(V, -1));
   int res = dfs(0, 0);
   cout << (res < INF ? res : -1) << endl;
}
```