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

int v,e;
int memo[16][100000];
vector<vector<pair<int,int> > > graph;

int dp(int n,int c,int bit){

  if(bit == (pow(2,v)-1)){
    for(int i = 0;i < graph[n].size();i++){
      if(graph[n][i].first == 0){
        return graph[n][i].second;
      }
    }
    return 1000000;
  }

  if(memo[n][bit] != 0){
    return memo[n][bit];
  }

  int a = 1000000;

  for(int i = 0;i < graph[n].size();i++){
    if((bit & (1 << graph[n][i].first)) == 0){
      a = min(a,dp(graph[n][i].first,c+1,bit | (1 << graph[n][i].first))+graph[n][i].second);
    }
  }

  return memo[n][bit] = a;
}



int main(){
  cin >> v >> e;
  graph.resize(v);
  for(int i = 0;i < e;i++){
    int s,t,d;
    cin >> s >> t >> d;
    pair<int,int> p = make_pair(t,d);
    graph[s].push_back(p);
  }

  int ans = dp(0,0,1);

  if(ans >= 20000){
    cout << -1 << endl;
  }else{
    cout << ans << endl;
  }

  return 0;
}


### Prompt

Please create a solution in CPP 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;

int v,e;
int memo[16][100000];
vector<vector<pair<int,int> > > graph;

int dp(int n,int c,int bit){

  if(bit == (pow(2,v)-1)){
    for(int i = 0;i < graph[n].size();i++){
      if(graph[n][i].first == 0){
        return graph[n][i].second;
      }
    }
    return 1000000;
  }

  if(memo[n][bit] != 0){
    return memo[n][bit];
  }

  int a = 1000000;

  for(int i = 0;i < graph[n].size();i++){
    if((bit & (1 << graph[n][i].first)) == 0){
      a = min(a,dp(graph[n][i].first,c+1,bit | (1 << graph[n][i].first))+graph[n][i].second);
    }
  }

  return memo[n][bit] = a;
}



int main(){
  cin >> v >> e;
  graph.resize(v);
  for(int i = 0;i < e;i++){
    int s,t,d;
    cin >> s >> t >> d;
    pair<int,int> p = make_pair(t,d);
    graph[s].push_back(p);
  }

  int ans = dp(0,0,1);

  if(ans >= 20000){
    cout << -1 << endl;
  }else{
    cout << ans << endl;
  }

  return 0;
}


```