#include <iostream>

using namespace std;

int main(){
    int INF = 1000000000;

    int N, M;
    cin >> N >> M;
    int g[N][N];

    for(int i=0; i<N; i++){
        for(int j=0; j<N; j++){
            g[i][j] = -1;
        }
    }
    for(int i=0; i<M; i++){
        int s, t, d;
        cin >> s >> t >> d;
        g[s][t] = d;
    }

    int dp[(1<<N)][N];
    for(int i=0; i<(1<<N); i++){
        for(int j=0; j<N; j++){
            dp[i][j] = INF;
        }
    }
    dp[0][0] = 0;
    for(int S=1; S<=(1<<N)-1; S++){
        for(int i=0; i<N; i++){
            for(int j=0; j<N; j++){
                int t = S ^ (1<<j);
                if(g[i][j] == -1) continue;
                if(!(S>>j)) continue;
                if(dp[t][i] == INF) continue;
                dp[S][j] = min(dp[S][j], dp[t][i]+g[i][j]);
            }
        }
    }

    if(dp[(1<<N)-1][0] == INF) cout << -1 << endl;
    else cout << dp[(1<<N)-1][0] <<endl;
    
    return 0;
}

### 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 <iostream>

using namespace std;

int main(){
    int INF = 1000000000;

    int N, M;
    cin >> N >> M;
    int g[N][N];

    for(int i=0; i<N; i++){
        for(int j=0; j<N; j++){
            g[i][j] = -1;
        }
    }
    for(int i=0; i<M; i++){
        int s, t, d;
        cin >> s >> t >> d;
        g[s][t] = d;
    }

    int dp[(1<<N)][N];
    for(int i=0; i<(1<<N); i++){
        for(int j=0; j<N; j++){
            dp[i][j] = INF;
        }
    }
    dp[0][0] = 0;
    for(int S=1; S<=(1<<N)-1; S++){
        for(int i=0; i<N; i++){
            for(int j=0; j<N; j++){
                int t = S ^ (1<<j);
                if(g[i][j] == -1) continue;
                if(!(S>>j)) continue;
                if(dp[t][i] == INF) continue;
                dp[S][j] = min(dp[S][j], dp[t][i]+g[i][j]);
            }
        }
    }

    if(dp[(1<<N)-1][0] == INF) cout << -1 << endl;
    else cout << dp[(1<<N)-1][0] <<endl;
    
    return 0;
}

```