#include <bits/stdc++.h>
#define N 15
using namespace std;
typedef pair<int,int> P;
int n,G[N][N];
int Min(int &a,int b){return a=min(a,b);}

int mem[N][1<<N],used[N][1<<N];
int dfs(int pos,int bit,int start){
  if(bit == (1<<n)-1) return 0;
  if(bit!=0 && pos == start) return 1e9;
  if(used[pos][bit]++) return mem[pos][bit];
  int res = 1e9;
  for(int i=0;i<n;i++){
    if(bit>>i&1)continue;
    int nbit = bit | 1<<i;
    Min(res,G[pos][i]+dfs(i,nbit,start));
  }
  return mem[pos][bit] = res;
}

int main(){
  int m;
  cin>>n>>m;

  for(int i=0;i<N;i++)
    for(int j=0;j<N;j++)G[i][j] = 1e9;
  
  for(int i=0,a,b,d;i<m;i++){
    cin>>a>>b>>d;
    Min(G[a][b],d);
  }

  int ans = 1e9;
  for(int i=0;i<n;i++){
    memset(used,0,sizeof(used));
    Min(ans,dfs(i,0,i));
  }
  if(ans == 1e9)cout<<-1<<endl;
  else cout<<ans<<endl;
  
  return 0;
}
### Prompt

In CPP, your task is to solve 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 N 15
using namespace std;
typedef pair<int,int> P;
int n,G[N][N];
int Min(int &a,int b){return a=min(a,b);}

int mem[N][1<<N],used[N][1<<N];
int dfs(int pos,int bit,int start){
  if(bit == (1<<n)-1) return 0;
  if(bit!=0 && pos == start) return 1e9;
  if(used[pos][bit]++) return mem[pos][bit];
  int res = 1e9;
  for(int i=0;i<n;i++){
    if(bit>>i&1)continue;
    int nbit = bit | 1<<i;
    Min(res,G[pos][i]+dfs(i,nbit,start));
  }
  return mem[pos][bit] = res;
}

int main(){
  int m;
  cin>>n>>m;

  for(int i=0;i<N;i++)
    for(int j=0;j<N;j++)G[i][j] = 1e9;
  
  for(int i=0,a,b,d;i<m;i++){
    cin>>a>>b>>d;
    Min(G[a][b],d);
  }

  int ans = 1e9;
  for(int i=0;i<n;i++){
    memset(used,0,sizeof(used));
    Min(ans,dfs(i,0,i));
  }
  if(ans == 1e9)cout<<-1<<endl;
  else cout<<ans<<endl;
  
  return 0;
}
```