#include <bits/stdc++.h>

using namespace std;

#define MAX_V 15
#define INF 1e9

int V,E;
int d[MAX_V][MAX_V];
int dp[1<<MAX_V][MAX_V];

int rec(int S,int v){
  if(dp[S][v] >= 0){
    return dp[S][v];
  }
  if(S == (1<<V)-1 && v == 0){
    return dp[S][v] = 0;
  }
  int res = INF;
  for(int u = 0 ; u < V ; u++){
    if(!(S >> u & 1)){
      res = min(res,rec(S|1<<u,u) + d[v][u]);
    }
  }
  return dp[S][v] = res;
}

void init(){
  memset(dp,-1,sizeof(dp));
  fill(d[0],d[0]+MAX_V*MAX_V,INF);
}

int main(){
  init();
  cin >> V >> E;
  for(int i = 0 ; i < E ; i++){
    int a,b,c;
    cin >> a >> b >> c;
    d[a][b] = c;
  }
  int res = rec(0,0);
  cout << (res == INF ? -1 : res) << endl;
  return 0;
}
### Prompt

Your challenge is to write 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 <bits/stdc++.h>

using namespace std;

#define MAX_V 15
#define INF 1e9

int V,E;
int d[MAX_V][MAX_V];
int dp[1<<MAX_V][MAX_V];

int rec(int S,int v){
  if(dp[S][v] >= 0){
    return dp[S][v];
  }
  if(S == (1<<V)-1 && v == 0){
    return dp[S][v] = 0;
  }
  int res = INF;
  for(int u = 0 ; u < V ; u++){
    if(!(S >> u & 1)){
      res = min(res,rec(S|1<<u,u) + d[v][u]);
    }
  }
  return dp[S][v] = res;
}

void init(){
  memset(dp,-1,sizeof(dp));
  fill(d[0],d[0]+MAX_V*MAX_V,INF);
}

int main(){
  init();
  cin >> V >> E;
  for(int i = 0 ; i < E ; i++){
    int a,b,c;
    cin >> a >> b >> c;
    d[a][b] = c;
  }
  int res = rec(0,0);
  cout << (res == INF ? -1 : res) << endl;
  return 0;
}
```