#include<math.h>
#include<cmath>
#include<bitset>
#include<stdio.h>
#include<string>
#include<map>
#include<algorithm>
#include<vector>
#include<iostream>
#include<utility>
using namespace std;
int main(){
  int INF=(1<<29);
  int v,e,s,t,d,g[20][20],i,j,k,dp[(1<<16)][20];
  scanf("%d %d\n",&v,&e);
  for(i=0;i<v;i++){
    for(j=0;j<v;j++){
      g[i][j]=-1;
    }
  }
  for(i=0;i<e;i++){
    scanf("%d %d %d\n",&s,&t,&d);
    g[s][t]=d;
  }
  for(i=0;i<(1<<v);i++){
    for(j=0;j<v;j++){
      dp[i][j]=INF;
    }
  }
  dp[0][0]=0;
  for(i=0;i<(1<<v);i++){
    for(j=0;j<v;j++){
      for(k=0;k<v;k++){
        if(g[j][k]!=-1 && (i&(1<<k))==0){
          dp[i|(1<<k)][k]=min(dp[i][j]+g[j][k],dp[i|(1<<k)][k]);
        }
      }
    }
  }
  if(dp[(1<<v)-1][0]==INF){
    printf("-1\n");
  }
  else{
    printf("%d\n",dp[(1<<v)-1][0]);
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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<math.h>
#include<cmath>
#include<bitset>
#include<stdio.h>
#include<string>
#include<map>
#include<algorithm>
#include<vector>
#include<iostream>
#include<utility>
using namespace std;
int main(){
  int INF=(1<<29);
  int v,e,s,t,d,g[20][20],i,j,k,dp[(1<<16)][20];
  scanf("%d %d\n",&v,&e);
  for(i=0;i<v;i++){
    for(j=0;j<v;j++){
      g[i][j]=-1;
    }
  }
  for(i=0;i<e;i++){
    scanf("%d %d %d\n",&s,&t,&d);
    g[s][t]=d;
  }
  for(i=0;i<(1<<v);i++){
    for(j=0;j<v;j++){
      dp[i][j]=INF;
    }
  }
  dp[0][0]=0;
  for(i=0;i<(1<<v);i++){
    for(j=0;j<v;j++){
      for(k=0;k<v;k++){
        if(g[j][k]!=-1 && (i&(1<<k))==0){
          dp[i|(1<<k)][k]=min(dp[i][j]+g[j][k],dp[i|(1<<k)][k]);
        }
      }
    }
  }
  if(dp[(1<<v)-1][0]==INF){
    printf("-1\n");
  }
  else{
    printf("%d\n",dp[(1<<v)-1][0]);
  }
  return 0;
}

```