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 d8x[] = {-1, -1, 0, 1, 1, 1, 0, -1};
int d8y[] = {0, 1, 1, 1, 0, -1, -1, -1};
struct info {
  int Ai, Bi, Ci;
  bool operator<(const info& p) const {
    if (Ci != p.Ci) return Ci < p.Ci;
    return Bi < p.Bi;
  }
} Q[1000007];
bool mark[1000007];
int quali[1000007];
int main(void) {
  int num;
  scanf("%d", &num);
  for (int i = 1; i <= num; i++) scanf("%d", &quali[i]);
  int app;
  scanf("%d", &app);
  for (int i = 0; i < app; i++) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    Q[i].Ai = x;
    Q[i].Bi = y;
    Q[i].Ci = z;
  }
  sort(Q, Q + app);
  long long ans = 0;
  int counter = 0;
  for (int i = 0; i < app; i++) {
    int value = Q[i].Ci;
    int a = Q[i].Ai;
    int b = Q[i].Bi;
    if (!mark[b] && quali[a] > quali[b]) {
      mark[b] = true;
      ans += value;
      counter++;
      if (counter == num - 1) break;
    }
  }
  if (counter == num - 1)
    printf("%lld\n", ans);
  else
    printf("-1");
  return 0;
}
