Nick's company employed n people. Now Nick needs to build a tree hierarchy of «supervisor-surbodinate» relations in the company (this is to say that each employee, except one, has exactly one supervisor). There are m applications written in the following form: «employee ai is ready to become a supervisor of employee bi at extra cost ci». The qualification qj of each employee is known, and for each application the following is true: qai > qbi. 

Would you help Nick calculate the minimum cost of such a hierarchy, or find out that it is impossible to build it.

Input

The first input line contains integer n (1 ≤ n ≤ 1000) — amount of employees in the company. The following line contains n space-separated numbers qj (0 ≤ qj ≤ 106)— the employees' qualifications. The following line contains number m (0 ≤ m ≤ 10000) — amount of received applications. The following m lines contain the applications themselves, each of them in the form of three space-separated numbers: ai, bi and ci (1 ≤ ai, bi ≤ n, 0 ≤ ci ≤ 106). Different applications can be similar, i.e. they can come from one and the same employee who offered to become a supervisor of the same person but at a different cost. For each application qai > qbi.

Output

Output the only line — the minimum cost of building such a hierarchy, or -1 if it is impossible to build it.

Examples

Input

4
7 2 3 1
4
1 2 5
2 4 1
3 4 1
1 3 5


Output

11


Input

3
1 2 3
2
3 1 2
3 1 3


Output

-1

Note

In the first sample one of the possible ways for building a hierarchy is to take applications with indexes 1, 2 and 4, which give 11 as the minimum total cost. In the second sample it is impossible to build the required hierarchy, so the answer is -1.
#include <bits/stdc++.h>
using namespace std;
long long get_parent(vector<int> &parent, int a) {
  if (parent[a] == a)
    return a;
  else {
    parent[a] = get_parent(parent, parent[a]);
    return parent[a];
  }
}
void union_set(vector<int> &parent, int a, int b) {
  a = get_parent(parent, a);
  b = get_parent(parent, b);
  parent[b] = a;
}
int main() {
  int n;
  scanf("%d", &n);
  int q;
  for (int i = 0; i < n; i++) {
    scanf("%d", &q);
  }
  int m, count = 0;
  scanf("%d", &m);
  vector<int> parent(n + 1);
  for (int i = 1; i <= n; i++) parent[i] = i;
  vector<vector<long long> > edges;
  long long u, v, w, res = 0;
  for (int i = 0; i < m; i++) {
    cin >> u >> v >> w;
    edges.push_back({w, u, v});
  }
  sort(edges.begin(), edges.end());
  for (auto e : edges) {
    int a = e[1], b = e[2], cost = e[0];
    if (get_parent(parent, a) != get_parent(parent, b) &&
        get_parent(parent, b) == b) {
      union_set(parent, a, b);
      res += cost;
      count++;
    }
  }
  if (count != n - 1) {
    cout << "-1\n";
  } else {
    printf("%lld\n", res);
  }
}
