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;
const int N = 1010;
struct App {
  int a, b, cost;
  App(int a, int b, int cost) : a(a), b(b), cost(cost) {}
  bool operator<(const App& o) const { return cost < o.cost; }
};
int calc(int n, vector<App>& apps) {
  int pa[N];
  fill_n(pa, N, -1);
  sort(apps.begin(), apps.end());
  int cost = 0;
  for (int i = 0; i < (int)(apps.size()); i++) {
    if (pa[apps[i].b] != -1) continue;
    cost += apps[i].cost;
    pa[apps[i].b] = apps[i].a;
  }
  int root = count(pa + 1, pa + n + 1, -1);
  if (root > 1) return -1;
  return cost;
}
int main() {
  int n;
  scanf("%d", &n);
  vector<int> quals(n);
  for (int i = 0; i < (int)(n); i++) scanf("%d", &quals[i]);
  int m;
  scanf("%d", &m);
  vector<App> apps;
  for (int i = 0; i < (int)(m); i++) {
    int a, b, c;
    scanf("%d %d %d", &a, &b, &c);
    apps.push_back(App(a, b, c));
  }
  cout << (calc(n, apps)) << '\n';
}
