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;
int dsu(int v, vector<int>& king) {
  if (king[v] == v) {
    return v;
  } else {
    int newKing = dsu(king[v], king);
    king[v] = newKing;
    return king[v];
  }
}
int Kruskal(vector<pair<int, pair<int, int>>>& edges, vector<int>& king) {
  int ans = 0;
  for (int i = 0; i < edges.size(); i++) {
    int king1 = dsu(edges[i].second.first, king);
    int king2 = dsu(edges[i].second.second, king);
    if (king1 != king2 &&
        king[edges[i].second.second] == edges[i].second.second) {
      ans += edges[i].first;
      king[king2] = king1;
    }
  }
  for (int i = 0; i < king.size(); i++) {
    if (dsu(i, king) != dsu(0, king)) {
      return -1;
    }
  }
  return ans;
}
int main() {
  int n, q[100000], m, ans, a, b, c;
  vector<int> king;
  vector<pair<int, pair<int, int>>> edges;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> q[i];
    king.push_back(i);
  }
  cin >> m;
  for (int i = 0; i < m; i++) {
    cin >> a >> b >> c;
    edges.push_back({c, {--a, --b}});
  }
  sort(edges.begin(), edges.end());
  ans = Kruskal(edges, king);
  cout << ans;
}
