#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

int dp[1 << 15][15];
int n;
int m[15][15];

int rec(int s,int v){
  if(dp[s][v]>=0){
    return dp[s][v];
  }

  if(s ==((1 << n)-1) && v==0){
    //printf("v=0\n");

    return dp[s][v]=0;
  }

  int res = 1000*15+1;

  for(int u=0;u<n;u++){
    if((!(s >> u & 1)) && m[v][u]>=0){
      //printf("m:%d",m[v][u]);
      res = min(res,rec(s | (1 << u),u)+m[v][u]);
    } 
  }
  //printf("res=%d\n",res);
  return dp[s][v]=res;
}

int main(){
  int v,e;
  cin >> v >> e;
  n=v;
  
    memset(m,-1,sizeof(m));


  int s,t,d;
  for(int i=0;i<e;i++){
    cin >> s >> t >> d;
    m[s][t]=d;
  }
  
  memset(dp,-1,sizeof(dp));
  int ans=rec(0,0);
  if(ans==15001){
    printf("%d\n",-1);
  }
  else{
    printf("%d\n",ans);
  }
  
}

    
### Prompt

Generate 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 <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

int dp[1 << 15][15];
int n;
int m[15][15];

int rec(int s,int v){
  if(dp[s][v]>=0){
    return dp[s][v];
  }

  if(s ==((1 << n)-1) && v==0){
    //printf("v=0\n");

    return dp[s][v]=0;
  }

  int res = 1000*15+1;

  for(int u=0;u<n;u++){
    if((!(s >> u & 1)) && m[v][u]>=0){
      //printf("m:%d",m[v][u]);
      res = min(res,rec(s | (1 << u),u)+m[v][u]);
    } 
  }
  //printf("res=%d\n",res);
  return dp[s][v]=res;
}

int main(){
  int v,e;
  cin >> v >> e;
  n=v;
  
    memset(m,-1,sizeof(m));


  int s,t,d;
  for(int i=0;i<e;i++){
    cin >> s >> t >> d;
    m[s][t]=d;
  }
  
  memset(dp,-1,sizeof(dp));
  int ans=rec(0,0);
  if(ans==15001){
    printf("%d\n",-1);
  }
  else{
    printf("%d\n",ans);
  }
  
}

    
```