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

const int INF = 1LL << 60;

signed main(){
  cin.tie(0);
  ios::sync_with_stdio(false);

  int V,E; cin >> V >> E;
  vector< vector<int> > G(V,vector<int> (V,INF));
  for(int i = 0; i < E; i++){
    int s,t,d; cin >> s >> t >> d;
    G[s][t] = d;
  }

  vector< vector<int> > dp(1 << V,vector<int> (V,-1));
  function< int(int,int) > rec =
  [&](int S,int v){ //{すでに訪れた頂点,現在位置}
    if(dp[S][v] >= 0) return dp[S][v];
    if(S == (1 << V)-1 && v == 0){
      //全ての頂点を訪れて戻ってきた
      return dp[S][v] = 0;
    }

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

  int ans = rec(0,0);
  if(ans == INF) cout << -1 << endl;
  else cout << ans << 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 int long long

const int INF = 1LL << 60;

signed main(){
  cin.tie(0);
  ios::sync_with_stdio(false);

  int V,E; cin >> V >> E;
  vector< vector<int> > G(V,vector<int> (V,INF));
  for(int i = 0; i < E; i++){
    int s,t,d; cin >> s >> t >> d;
    G[s][t] = d;
  }

  vector< vector<int> > dp(1 << V,vector<int> (V,-1));
  function< int(int,int) > rec =
  [&](int S,int v){ //{すでに訪れた頂点,現在位置}
    if(dp[S][v] >= 0) return dp[S][v];
    if(S == (1 << V)-1 && v == 0){
      //全ての頂点を訪れて戻ってきた
      return dp[S][v] = 0;
    }

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

  int ans = rec(0,0);
  if(ans == INF) cout << -1 << endl;
  else cout << ans << endl;

  return 0;
}


```