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>
int main() {
  long n;
  scanf("%ld", &n);
  for (long p = 0; p < n; p++) {
    long x;
    scanf("%ld", &x);
  }
  long m;
  scanf("%ld", &m);
  std::map<long, long> cost;
  while (m--) {
    long x, y, c;
    scanf("%ld %ld %ld", &x, &y, &c);
    if (cost.count(y) > 0) {
      cost[y] = (cost[y] < c) ? cost[y] : c;
    } else {
      cost[y] = c;
    }
  }
  if (cost.size() < n - 1) {
    puts("-1");
    return 0;
  }
  long long total(0);
  for (std::map<long, long>::iterator it = cost.begin(); it != cost.end();
       it++) {
    total += it->second;
  }
  printf("%lld\n", total);
  return 0;
}
