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 v, e;
int memo[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 memo[bit] = graph[n][i].second;
      }
    }
    return 1000000;
  }
  if (graph[n].size() == 0) {
    return 1000000;
  }
  int a = 20000;
  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[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;
}
