#include<iostream>
#define MAX_V 16
#define INF 100000000
using namespace std;
int n,s,t,l,m;
int dp[1<<MAX_V][MAX_V];
int d[MAX_V][MAX_V];
int solve(int S,int v){
  if(dp[S][v]>=0){
    return dp[S][v];
  }
  if(S==(1<<n)-1 && v==0){
    return dp[S][v]=0;
  }
  int res = INF;
  for(int u=0;u<n;u++){
    if(!(S>>u&1)){
      res = min(res,solve(S | 1<<u, u)+d[v][u]);
    }
  }
  return dp[S][v] = res;
}
int main(){
  cin >> n>>m;
  for(int i=0;i<n;i++){
    for(int j=0;j<n;j++){
      d[i][j]=INF;
    }
  }
  for(int i=0;i<1<<n;i++){
    for(int j=0;j<n;j++){
      dp[i][j]=-1;
    }
  }
  for(int i=0;i<m;i++){
    cin >> s >> t;
    cin >>d[s][t];
  }
  int tmp=solve(0,0);
  if(tmp==INF){
    cout << "-1\n";
  }else{
    cout << tmp<<endl;
  }
  
}


### Prompt

Construct a Cpp code solution to the problem outlined:

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<iostream>
#define MAX_V 16
#define INF 100000000
using namespace std;
int n,s,t,l,m;
int dp[1<<MAX_V][MAX_V];
int d[MAX_V][MAX_V];
int solve(int S,int v){
  if(dp[S][v]>=0){
    return dp[S][v];
  }
  if(S==(1<<n)-1 && v==0){
    return dp[S][v]=0;
  }
  int res = INF;
  for(int u=0;u<n;u++){
    if(!(S>>u&1)){
      res = min(res,solve(S | 1<<u, u)+d[v][u]);
    }
  }
  return dp[S][v] = res;
}
int main(){
  cin >> n>>m;
  for(int i=0;i<n;i++){
    for(int j=0;j<n;j++){
      d[i][j]=INF;
    }
  }
  for(int i=0;i<1<<n;i++){
    for(int j=0;j<n;j++){
      dp[i][j]=-1;
    }
  }
  for(int i=0;i<m;i++){
    cin >> s >> t;
    cin >>d[s][t];
  }
  int tmp=solve(0,0);
  if(tmp==INF){
    cout << "-1\n";
  }else{
    cout << tmp<<endl;
  }
  
}


```