#include<bits/stdc++.h>
#define r(i,n) for(int i=0;i<n;i++)
using namespace std;
#define MAX 15
#define INF 1<<28
int n;
int d[MAX][MAX];
int dp[1<<MAX][MAX];
//s=?¨??????????,v=?????¨??????
int rec(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,rec(s|1<<u,u)+d[v][u]);
  return dp[s][v]=res;
}
int main(){
  memset(dp,-1,sizeof(dp));
  r(i,MAX)r(j,MAX)d[i][j]=INF;
  int m;
  cin>>n>>m;
  while(m--){
    int s,t,dd;
    cin>>s>>t>>dd;
    d[s][t]=dd;
  }
  int p=rec(0,0);
  if(p==INF)p=-1;
  cout<<p<<endl;
}
### Prompt

Create a solution in Cpp for 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>
#define r(i,n) for(int i=0;i<n;i++)
using namespace std;
#define MAX 15
#define INF 1<<28
int n;
int d[MAX][MAX];
int dp[1<<MAX][MAX];
//s=?¨??????????,v=?????¨??????
int rec(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,rec(s|1<<u,u)+d[v][u]);
  return dp[s][v]=res;
}
int main(){
  memset(dp,-1,sizeof(dp));
  r(i,MAX)r(j,MAX)d[i][j]=INF;
  int m;
  cin>>n>>m;
  while(m--){
    int s,t,dd;
    cin>>s>>t>>dd;
    d[s][t]=dd;
  }
  int p=rec(0,0);
  if(p==INF)p=-1;
  cout<<p<<endl;
}
```