#include <bits/stdc++.h>
using namespace std;

constexpr int INF = 1e9;

int main() {
    int V, E;
    cin >> V >> E;
    vector<vector<int>> g(V, vector<int>(V, INF));
    for(int i=0; i<E; ++i) {
        int s, t, d;
        cin >> s >> t >> d;
        g[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 >> j & 1) {
                continue;
            }
            for(int k=0; k<V; ++k) {
                dp[j][i | (1 << j)] = min(dp[j][i | (1 << j)], dp[k][i] + g[k][j]);
            }
        }
    }
    if(dp[0][(1 << V) - 1] == INF) {
        cout << -1 << endl;
    } else {
        cout << dp[0][(1 << V) - 1] << endl;
    }
}

### Prompt

Your challenge is to write 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;

constexpr int INF = 1e9;

int main() {
    int V, E;
    cin >> V >> E;
    vector<vector<int>> g(V, vector<int>(V, INF));
    for(int i=0; i<E; ++i) {
        int s, t, d;
        cin >> s >> t >> d;
        g[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 >> j & 1) {
                continue;
            }
            for(int k=0; k<V; ++k) {
                dp[j][i | (1 << j)] = min(dp[j][i | (1 << j)], dp[k][i] + g[k][j]);
            }
        }
    }
    if(dp[0][(1 << V) - 1] == INF) {
        cout << -1 << endl;
    } else {
        cout << dp[0][(1 << V) - 1] << endl;
    }
}

```