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
#include <iostream>
#include <vector>
using namespace std;
const int INF = (1e9);

int main(){
  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;
  }
  for(int i = 0; i < V; ++i) G[i][i] = 0;
  vector< vector<int> > DP((1<<V),vector<int>(V,INF));
  DP[1][0] = 0;
  for(int i = 1; i < (1<<V); ++i){
    for(int j = 0; j < V; ++j){
      if(!((i>>j)&1)) continue;
      for(int k = 0; k < V; ++k){
        if(((i>>k) & 1) || j == k) continue;
        DP[i|(1<<k)][k] = min(DP[i|(1<<k)][k], DP[i][j] + G[j][k]);
        //cerr << "DP[" << (i|(1<<k)) << "] = " << DP[i|(1<<k)][k] << endl;
      }
    }
  }
  int ans = INF;
  for(int i = 0; i < V; ++i) ans = min(ans, DP[(1<<V)-1][i] + G[i][0]);
  if(ans < INF) cout << ans << endl;
  else cout << -1 << endl;
  return 0;
}

