#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i = 0; i < (n); i++)

const int MAX_V = 15;

int V, E;
int dist[MAX_V][MAX_V];
int dp[1<<MAX_V][MAX_V];
int s;
const int INF = 1000000;

int rec (int S, int v){
    if(dp[S][v] != -1) return dp[S][v];

    if(S == (1<<V)-1 && v == s) return dp[S][v] = 0;

    int res = INF;
    for(int u = 0; u < V; u++){
        if(!(S>>u & 1)){
            res = min(res, rec(S | (1<<u),u) + dist[v][u]);
        }
    }

    return dp[S][v] = res;
}

int main(){
    rep(i, MAX_V) rep(j, MAX_V) dist[i][j] = INF;
    cin >> V >> E;
    rep(i, E){
        int from, to, cost;
        cin >> from >> to >> cost;
        dist[from][to] = cost;
    }

    int res = INF;
    rep(i,V){
        memset (dp, -1, sizeof(dp));
        s = i;
        res = min(res, rec(0, i));
    }
    cout << (res == INF ? -1 : res) << 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;
#define rep(i,n) for(int i = 0; i < (n); i++)

const int MAX_V = 15;

int V, E;
int dist[MAX_V][MAX_V];
int dp[1<<MAX_V][MAX_V];
int s;
const int INF = 1000000;

int rec (int S, int v){
    if(dp[S][v] != -1) return dp[S][v];

    if(S == (1<<V)-1 && v == s) return dp[S][v] = 0;

    int res = INF;
    for(int u = 0; u < V; u++){
        if(!(S>>u & 1)){
            res = min(res, rec(S | (1<<u),u) + dist[v][u]);
        }
    }

    return dp[S][v] = res;
}

int main(){
    rep(i, MAX_V) rep(j, MAX_V) dist[i][j] = INF;
    cin >> V >> E;
    rep(i, E){
        int from, to, cost;
        cin >> from >> to >> cost;
        dist[from][to] = cost;
    }

    int res = INF;
    rep(i,V){
        memset (dp, -1, sizeof(dp));
        s = i;
        res = min(res, rec(0, i));
    }
    cout << (res == INF ? -1 : res) << endl;
}

```