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 main() {
  int n;
  scanf("%d", &n);
  vector<int> qs(n);
  int most = -1;
  bool more = false;
  for (int i = 0; i < n; i++) {
    scanf("%d", &qs[i]);
    if (qs[i] == most) {
      more = true;
    }
    if (qs[i] > most) {
      most = qs[i];
      more = false;
    }
  }
  if (more) {
    printf("-1\n");
    return 0;
  }
  int m;
  scanf("%d", &m);
  vector<int> cost(n, -1);
  for (int i = 0; i < m; i++) {
    int sup, sub, c;
    scanf("%d %d %d", &sup, &sub, &c);
    sup--;
    sub--;
    if (cost[sub] == -1 or c < cost[sub]) cost[sub] = c;
  }
  long long C = 0;
  for (int i = 0; i < n; i++) {
    if ((cost[i] == -1) and (qs[i] != most)) {
      printf("-1\n");
      return 0;
    }
    if (cost[i] != -1) {
      C += cost[i];
    }
  }
  printf("%lld\n", C);
}
