
#include <bits/stdc++.h>

using namespace std;

int dp[1 << 15][15];

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);

    int INF = 15000;
    int V, E;
    cin >> V >> E;
    
    int d[V][V];
    for (int i = 0; i < V; ++i) {
        fill(d[i], d[i] + V, INF);
    }
    for (int i = 0; i < E; ++i) {
        int s, t;
        cin >> s >> t;
        cin >> d[s][t];
    }
    for (int s = 0; s < 1 << V; ++s) {
        fill(dp[s], dp[s] + V, INF);
    }

    dp[(1 << V) - 1][0] = 0;

    for (int s = (1 << V) - 2; s >= 0; --s) {
        for (int v = 0; v < V; ++v) {
            for (int u = 0; u < V; ++u) {
                if (!(s >> u & 1)) {
                    dp[s][v] = min(dp[s][v], dp[s | 1 << u][u] + d[v][u]);
                }
            }
        }
    }

    if (dp[0][0] < INF) {
        cout << dp[0][0] << '\n';
    } else {
        cout << -1 << '\n';
    }

    return 0;
}


### Prompt

In Cpp, your task is to solve 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;

int dp[1 << 15][15];

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);

    int INF = 15000;
    int V, E;
    cin >> V >> E;
    
    int d[V][V];
    for (int i = 0; i < V; ++i) {
        fill(d[i], d[i] + V, INF);
    }
    for (int i = 0; i < E; ++i) {
        int s, t;
        cin >> s >> t;
        cin >> d[s][t];
    }
    for (int s = 0; s < 1 << V; ++s) {
        fill(dp[s], dp[s] + V, INF);
    }

    dp[(1 << V) - 1][0] = 0;

    for (int s = (1 << V) - 2; s >= 0; --s) {
        for (int v = 0; v < V; ++v) {
            for (int u = 0; u < V; ++u) {
                if (!(s >> u & 1)) {
                    dp[s][v] = min(dp[s][v], dp[s | 1 << u][u] + d[v][u]);
                }
            }
        }
    }

    if (dp[0][0] < INF) {
        cout << dp[0][0] << '\n';
    } else {
        cout << -1 << '\n';
    }

    return 0;
}


```