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 Max = 1e5 + 5;
int main() {
  int T = 1;
  while (T--) {
    vector<priority_queue<pair<int, int>, vector<pair<int, int>>,
                          greater<pair<int, int>>>>
        G;
    int n;
    scanf("%d", &n);
    G.resize(n + 1);
    int q[1005];
    for (int i = 1; i <= n; i++) {
      scanf("%d", &q[i]);
    }
    int m;
    scanf("%d", &m);
    int u, v, w;
    for (int i = 0; i < m; i++) {
      scanf("%d%d%d", &u, &v, &w);
      if (q[u] > q[v]) {
        G[v].push(make_pair(w, u));
      }
    }
    int edges = 0;
    long long ans = 0;
    for (int i = 1; i <= n && edges < n - 1; i++) {
      if (!G[i].empty()) {
        ans += G[i].top().first;
        edges++;
      }
    }
    if (edges != n - 1)
      printf("-1");
    else
      printf("%lld", ans);
  }
  return 0;
}
