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 <bits/stdc++.h>
using namespace std;
class Solver {
 private:
  int v;
  int e;
  vector<vector<int> > adj;
  vector<vector<int> > dp;

 public:
  Solver() {
    cin >> v >> e;
    adj.resize(v);
    for (int i = 0; i < v; ++i) {
      adj[i].resize(v);
      for (int j = 0; j < v; ++j) {
        adj[i][j] = INT_MAX / 2;
      }
    }
    for (int i = 0; i < e; ++i) {
      int s, t, d;
      cin >> s >> t >> d;
      adj[s][t] = d;
    }
    dp = vector<vector<int> >(1 << v, vector<int>(v, INT_MAX / 2));
  }
  int recursive(int bit, int cur) {
    if (dp[bit][cur] != INT_MAX / 2) {
      return dp[bit][cur];
    }
    if (bit == (1 << v) - 1 && cur == 0) return dp[bit][cur] = 0;
    int ans = INT_MAX / 2;
    for (int i = 0; i < v; ++i) {
      if (!(bit >> i & 1) && adj[cur][i] != INT_MAX / 2) {
        if (recursive(bit | 1 << i, i) == INT_MAX / 2) continue;
        ans = min(ans, recursive(bit | 1 << i, i) + adj[cur][i]);
      }
    }
    return dp[bit][cur] = ans;
  }
  void exec() {
    int ans = recursive(0, 0);
    if (ans == INT_MAX / 2) ans = -1;
    cout << ans << endl;
  }
};
int main() {
  Solver s = Solver();
  s.exec();
}
