#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int IINF = INT_MAX;
#define REP(i, n) for (int i = 0; i < n; i++)

int V, E, d[20][20], dp[15][1<<15];

int main() {
    cin >> V >> E;
    REP(i,V) fill(d[i], d[i]+V, IINF/3);
    REP(i,E){
        int s, t;
        cin >> s >> t;
        cin >> d[s][t];
    }
    REP(i,V) fill(dp[i], dp[i]+(1<<V), IINF/3);
    dp[0][0] = 0;
    REP(S,(1<<V)){
        REP(i,V){
            REP(j,V){
                if(((S>>j)&1)==0){
                    dp[j][S|(1<<j)] = min(dp[j][S|(1<<j)], dp[i][S]+d[i][j]);
                }
            }
        }
    }
    cout << (dp[0][(1<<V)-1]<IINF/3?dp[0][(1<<V)-1]:-1) << endl;
    return 0;
}

### 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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int IINF = INT_MAX;
#define REP(i, n) for (int i = 0; i < n; i++)

int V, E, d[20][20], dp[15][1<<15];

int main() {
    cin >> V >> E;
    REP(i,V) fill(d[i], d[i]+V, IINF/3);
    REP(i,E){
        int s, t;
        cin >> s >> t;
        cin >> d[s][t];
    }
    REP(i,V) fill(dp[i], dp[i]+(1<<V), IINF/3);
    dp[0][0] = 0;
    REP(S,(1<<V)){
        REP(i,V){
            REP(j,V){
                if(((S>>j)&1)==0){
                    dp[j][S|(1<<j)] = min(dp[j][S|(1<<j)], dp[i][S]+d[i][j]);
                }
            }
        }
    }
    cout << (dp[0][(1<<V)-1]<IINF/3?dp[0][(1<<V)-1]:-1) << endl;
    return 0;
}

```