#include<bits/stdc++.h>
using namespace std;

typedef pair<int,int> P;
typedef vector<P>Vec;

int v,e;
Vec Edge[15];
int memo[1<<16][15];
int solve(int bit,int now);

int main(){
  memset(memo,-1,sizeof(memo));
  int s,t,d;
  cin>>v>>e;
  for(int i=0;i<e;i++){
    cin>>s>>t>>d;
    Edge[s].push_back(P(t,d));
  }
  int ans=solve(0,0);
  if(ans==(1<<30))cout<<"-1"<<endl;
  else cout<<ans<<endl;
}

int solve(int bit,int now){
  int cost=1<<30;
  if(memo[bit][now]!=-1)return memo[bit][now];
  // ->goal
  if(bit== (1<<v)-1)return 0;
  else if(bit & 1)return cost;
  // <-goal
  
  for(int i=0;i<Edge[now].size();i++){
if((bit & 1 << Edge[now][i].first)==0){
      cost=min(cost,solve(bit | 1 << Edge[now][i].first,Edge[now][i].first)+Edge[now][i].second);
    }
  } 
  return memo[bit][now]=cost;
}
### 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;

typedef pair<int,int> P;
typedef vector<P>Vec;

int v,e;
Vec Edge[15];
int memo[1<<16][15];
int solve(int bit,int now);

int main(){
  memset(memo,-1,sizeof(memo));
  int s,t,d;
  cin>>v>>e;
  for(int i=0;i<e;i++){
    cin>>s>>t>>d;
    Edge[s].push_back(P(t,d));
  }
  int ans=solve(0,0);
  if(ans==(1<<30))cout<<"-1"<<endl;
  else cout<<ans<<endl;
}

int solve(int bit,int now){
  int cost=1<<30;
  if(memo[bit][now]!=-1)return memo[bit][now];
  // ->goal
  if(bit== (1<<v)-1)return 0;
  else if(bit & 1)return cost;
  // <-goal
  
  for(int i=0;i<Edge[now].size();i++){
if((bit & 1 << Edge[now][i].first)==0){
      cost=min(cost,solve(bit | 1 << Edge[now][i].first,Edge[now][i].first)+Edge[now][i].second);
    }
  } 
  return memo[bit][now]=cost;
}
```