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;
long long INF = 1e18;
long long rec(int S, int v, const int V, vector<vector<long long>>& dp,
              const vector<vector<long long>>& dist) {
  if (dp[S][v] >= 0) return dp[S][v];
  if (S == (1 << V) - 1 && v == 0) {
    return dp[S][v] = 0;
  }
  long long res = INF;
  for (int u = 0; u < (int)(V); ++u) {
    if (!(S | 1 << u)) {
      auto d = dist[v][u];
      if (d != INF) {
        res = min(res, rec(S | 1 << u, u, V, dp, dist) + d);
      }
    }
  }
  return dp[S][v] = res;
}
long long travelling_salesman(int V, vector<vector<long long>>& dist) {
  vector<vector<long long>> dp(1 << V, vector<long long>(V, -INF));
  return rec(0, 0, V, dp, dist);
}
int main() {
  int V, E;
  cin >> V >> E;
  vector<vector<long long>> dist(V, vector<long long>(V, INF));
  for (int e = 0; e < (int)(E); ++e) {
    int s, t;
    long long d;
    cin >> s >> t >> d;
    dist[s][t] = d;
  }
  auto ret = travelling_salesman(V, dist);
  cout << (ret == INF ? -1 : ret) << endl;
}
