#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<n;++i)
const int MAX_V = 15;
const int INF = 1e9;
 
int V,E;
int d[MAX_V][MAX_V];
int dp[MAX_V][1<<MAX_V];

int dfs(int v, int mask){
    if(mask==(1<<V)-1) return d[v][0];
    if(dp[v][mask]!=-1) return dp[v][mask];
    int res = INF;
    rep(u,V){
        if(v==u || d[v][u]==INF || (mask&(1<<u))) continue;
        res = min(res, dfs(u, mask|(1<<u))+d[v][u]);
    }
    return dp[v][mask] = res;
}

int main() {
    cin>>V>>E;
    rep(i,V)rep(j,V) d[i][j]=INF;
    rep(i,V) d[i][i]=0;
    rep(i,E){
        int s,t;
        cin>>s>>t;
        cin>>d[s][t];
    }
    memset(dp, -1, sizeof(dp));
    int res=dfs(0,1);
    cout<<(res==INF ? -1 : res)<<endl;
    return 0;
}
### Prompt

Your challenge is to write a cpp solution 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;
#define rep(i,n) for(int i=0;i<n;++i)
const int MAX_V = 15;
const int INF = 1e9;
 
int V,E;
int d[MAX_V][MAX_V];
int dp[MAX_V][1<<MAX_V];

int dfs(int v, int mask){
    if(mask==(1<<V)-1) return d[v][0];
    if(dp[v][mask]!=-1) return dp[v][mask];
    int res = INF;
    rep(u,V){
        if(v==u || d[v][u]==INF || (mask&(1<<u))) continue;
        res = min(res, dfs(u, mask|(1<<u))+d[v][u]);
    }
    return dp[v][mask] = res;
}

int main() {
    cin>>V>>E;
    rep(i,V)rep(j,V) d[i][j]=INF;
    rep(i,V) d[i][i]=0;
    rep(i,E){
        int s,t;
        cin>>s>>t;
        cin>>d[s][t];
    }
    memset(dp, -1, sizeof(dp));
    int res=dfs(0,1);
    cout<<(res==INF ? -1 : res)<<endl;
    return 0;
}
```