#include<bits/stdc++.h>
using namespace std;
#define INF 1e9+1
int n,d[20][20];
int dp[1<<20][20];

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 i=0;i<n;i++){
    if((!(s>>i&1))&&(d[v][i]>=0))
      res=min(res,rec(s|1<<i,i)+d[v][i]);
  }
  return dp[s][v]=res;
}

int main(){
  int e,s,t,l;
  cin>>n>>e;
  memset(dp,-1,sizeof(dp));
  memset(d,-1,sizeof(d));
  for(int i=0;i<e;i++){
    cin>>s>>t>>l;
    d[s][t]=l;
  }
  int ans=rec(0,0);
  cout<<(ans==INF?-1:ans)<<endl;
  return 0;
}


### 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<bits/stdc++.h>
using namespace std;
#define INF 1e9+1
int n,d[20][20];
int dp[1<<20][20];

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 i=0;i<n;i++){
    if((!(s>>i&1))&&(d[v][i]>=0))
      res=min(res,rec(s|1<<i,i)+d[v][i]);
  }
  return dp[s][v]=res;
}

int main(){
  int e,s,t,l;
  cin>>n>>e;
  memset(dp,-1,sizeof(dp));
  memset(d,-1,sizeof(d));
  for(int i=0;i<e;i++){
    cin>>s>>t>>l;
    d[s][t]=l;
  }
  int ans=rec(0,0);
  cout<<(ans==INF?-1:ans)<<endl;
  return 0;
}


```