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;
int main() {
  int v, e;
  cin >> v >> e;
  int graph[15][15];
  for (int i = 0; i < v; i++) {
    for (int j = 0; j < v; j++) {
      graph[i][j] = 1000000007;
    }
  }
  for (int i = 0; i < e; i++) {
    int s, t, d;
    cin >> s >> t >> d;
    graph[s][t] = d;
  }
  pair<int, int> result[(1 << 15)][15];
  for (int i = 0; i < (1 << v); i++) {
    for (int j = 0; j < v; j++) {
      result[i][j] = make_pair(1000000007, -1);
    }
  }
  for (int i = 0; i < v; i++) {
    result[(1 << i)][i] = make_pair(0, i);
  }
  for (int i = 1; i < (1 << v); i++) {
    for (int j = 0; j < v; j++) {
      if (((1 << j) & i) == 0) continue;
      int tmp = (i ^ (1 << j));
      for (int k = 0; k < v; k++) {
        if (((1 << k) & tmp) == 0) continue;
        if (result[i][j].first > result[tmp][k].first + graph[k][j]) {
          result[i][j].first = result[tmp][k].first + graph[k][j];
          result[i][j].second = result[tmp][k].second;
        }
      }
    }
  }
  int ans = 1000000007;
  for (int i = 0; i < v; i++) {
    if (result[(1 << v) - 1][i].first == 1000000007) continue;
    ans = min(ans, result[(1 << v) - 1][i].first +
                       graph[i][result[(1 << v) - 1][i].second]);
  }
  if (ans == 1000000007)
    cout << -1 << endl;
  else
    cout << ans << endl;
  return 0;
}
