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

#define REP(i, n) for(int i = 0; i < n; i++)
#define REPR(i, n) for(int i = n; i >= 0; i--)
#define FOR(i, m, n) for(int i = m; i < n; i++)
#define ALL(obj) (obj).begin(), (obj).end()
#define INF 1e9
typedef long long ll;

int V,E;


vector<vector<int>> G(15,vector<int>(15,INF));

vector<vector<int>> dp((1<<15), vector<int>(15,-1));

int rec(int bit, int v){

    if (dp[bit][v] >= 0) return dp[bit][v];//メモ化

    if(bit==0){//初期化
        if(v==0){
            return 0;
        }else{
            return INF;
        }
    }

    if((bit&(1<<v)) == 0) return INF;//、矛盾する状態なら

    dp[bit][v] = INF;//ここからdp

    REP(i,V){
        dp[bit][v] = min(dp[bit][v], rec(bit^(1<<v),i) + G[i][v]);
    }
    return dp[bit][v];
}

int main() {
    cin >> V >> E;
    REP(i, E){
        int s,t,d;
        cin >> s >> t >> d;
        G[s][t] = d;
    }

    int ans = rec((1<<V)-1, 0);

    if(ans == INF) ans = -1;

    cout << ans << endl;

    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;

#define REP(i, n) for(int i = 0; i < n; i++)
#define REPR(i, n) for(int i = n; i >= 0; i--)
#define FOR(i, m, n) for(int i = m; i < n; i++)
#define ALL(obj) (obj).begin(), (obj).end()
#define INF 1e9
typedef long long ll;

int V,E;


vector<vector<int>> G(15,vector<int>(15,INF));

vector<vector<int>> dp((1<<15), vector<int>(15,-1));

int rec(int bit, int v){

    if (dp[bit][v] >= 0) return dp[bit][v];//メモ化

    if(bit==0){//初期化
        if(v==0){
            return 0;
        }else{
            return INF;
        }
    }

    if((bit&(1<<v)) == 0) return INF;//、矛盾する状態なら

    dp[bit][v] = INF;//ここからdp

    REP(i,V){
        dp[bit][v] = min(dp[bit][v], rec(bit^(1<<v),i) + G[i][v]);
    }
    return dp[bit][v];
}

int main() {
    cin >> V >> E;
    REP(i, E){
        int s,t,d;
        cin >> s >> t >> d;
        G[s][t] = d;
    }

    int ans = rec((1<<V)-1, 0);

    if(ans == INF) ans = -1;

    cout << ans << endl;

    return 0;
}

```