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 N, M;
int Q[1000];
pair<int, pair<int, int> > app[10000];
bool marked[1000];
int solve() {
  sort(app, app + M);
  int root = 0;
  bool bad = false;
  for (int i = 1; i < N; i++) {
    if (Q[root] < Q[i]) {
      root = i;
      bad = false;
    } else if (Q[root] == Q[i]) {
      bad = true;
    }
  }
  if (bad) {
    return -1;
  }
  int cost = 0;
  fill(marked, marked + N, false);
  marked[root] = true;
  int X = 1;
  for (int i = 0; i < N; i++) {
    if (i != root) {
      bool found = false;
      for (int j = 0; j < M; j++) {
        if (app[j].second.second == i) {
          cost += app[j].first;
          found = true;
          break;
        }
      }
      if (!found) {
        return -1;
      }
    }
  }
  return cost;
}
inline void init() {}
int main() {
  init();
  while (cin >> N) {
    for (int i = 0; i < N; i++) {
      cin >> Q[i];
    }
    cin >> M;
    for (int i = 0; i < M; i++) {
      cin >> app[i].second.first >> app[i].second.second >> app[i].first;
      app[i].second.first--;
      app[i].second.second--;
    }
    cout << solve() << endl;
  }
  return 0;
}
