#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define ALL(v) v.begin(), v.end()
typedef long long ll;

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

const int INF=1e9;

int main(){
  int v,e;
  cin>>v>>e;
  
  int G[15][15];
  rep(i,v){
    rep(j,v) G[i][j]=INF;
  }
  
  int s,t,d;
  rep(i,e){
    cin>>s>>t>>d;
    G[s][t]=d;
  }
  
  int dp[1<<v][v];
  rep(i,(1<<v)){
    rep(j,v) dp[i][j]=INF;
  }
  dp[0][0]=0;
  
  for(int s=0;s<(1<<v);s++){
    for(int i=0;i<v;i++){
      for(int j=0;j<v;j++){
        if((s&(1<<j))==0) continue;
        if(G[i][j]==INF) continue;
        dp[s][j]=min(dp[s][j],dp[s-(1<<j)][i]+G[i][j]);
      }
    }
  }
  
  if(dp[(1<<v)-1][0]==INF) cout<<-1<<endl;
  else cout<<dp[(1<<v)-1][0]<<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
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define ALL(v) v.begin(), v.end()
typedef long long ll;

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

const int INF=1e9;

int main(){
  int v,e;
  cin>>v>>e;
  
  int G[15][15];
  rep(i,v){
    rep(j,v) G[i][j]=INF;
  }
  
  int s,t,d;
  rep(i,e){
    cin>>s>>t>>d;
    G[s][t]=d;
  }
  
  int dp[1<<v][v];
  rep(i,(1<<v)){
    rep(j,v) dp[i][j]=INF;
  }
  dp[0][0]=0;
  
  for(int s=0;s<(1<<v);s++){
    for(int i=0;i<v;i++){
      for(int j=0;j<v;j++){
        if((s&(1<<j))==0) continue;
        if(G[i][j]==INF) continue;
        dp[s][j]=min(dp[s][j],dp[s-(1<<j)][i]+G[i][j]);
      }
    }
  }
  
  if(dp[(1<<v)-1][0]==INF) cout<<-1<<endl;
  else cout<<dp[(1<<v)-1][0]<<endl;

  return 0;
}

```