#include <bits/stdc++.h>
using namespace std;
const int INF = 1145141919;
vector <vector<int>> d(15,vector <int>(15,-1));
vector <vector<int>> dp(1<<15,vector <int>(15,-1));
int V,E;
int rec(int S, int now){
    if(dp[S][now]!=-1) return dp[S][now];
    if(S==(1<<V)-1&&now==0) return 0;
    int r=INF;
    for(int next=0;next<V;next++){
        if(d[now][next]!=-1 && !(S & (1<<next)))
        r = min(r, rec(S|(1<<next), next)+d[now][next]);
    }
    dp[S][now]=r;
    return r;
}
int main(void){
    cin>>V>>E;
    for(int i=0;i<E;i++){
        int a,b; cin>>a>>b; cin>>d[a][b];
    }
    int ans = rec(0,0);
    if(ans==INF) cout<<-1<<endl;
    else cout<<ans<<endl;
}


### 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;
const int INF = 1145141919;
vector <vector<int>> d(15,vector <int>(15,-1));
vector <vector<int>> dp(1<<15,vector <int>(15,-1));
int V,E;
int rec(int S, int now){
    if(dp[S][now]!=-1) return dp[S][now];
    if(S==(1<<V)-1&&now==0) return 0;
    int r=INF;
    for(int next=0;next<V;next++){
        if(d[now][next]!=-1 && !(S & (1<<next)))
        r = min(r, rec(S|(1<<next), next)+d[now][next]);
    }
    dp[S][now]=r;
    return r;
}
int main(void){
    cin>>V>>E;
    for(int i=0;i<E;i++){
        int a,b; cin>>a>>b; cin>>d[a][b];
    }
    int ans = rec(0,0);
    if(ans==INF) cout<<-1<<endl;
    else cout<<ans<<endl;
}


```