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;
long long n, x, m, ans, fix[1001][1001], parent[10001], fixx[10001];
pair<long long, pair<int, int>> a[10001];
int P(int x) {
  if (parent[x] == x) return x;
  return parent[x] = P(parent[x]);
}
void dsu(int a, int b) {
  a = P(a);
  b = P(b);
  parent[a] = b;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> x, parent[i] = i;
  cin >> m;
  for (int i = 1; i <= m; i++) {
    cin >> a[i].second.first >> a[i].second.second >> a[i].first;
  }
  sort(a + 1, a + m + 1);
  for (int i = 1; i <= m; i++) {
    if (fixx[a[i].second.second]) continue;
    fixx[a[i].second.second] = 1;
    if (P(a[i].second.first) != P(a[i].second.second)) {
      dsu(a[i].second.first, a[i].second.second);
      ans += a[i].first;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (P(i) != P(1)) {
      cout << "-1";
      return 0;
    }
  }
  cout << ans;
}
