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 gcd(long long a, long long b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
int main() {
  int n;
  cin >> n;
  vector<int> q(n);
  for (int i = 0; i < n; ++i) cin >> q[i];
  vector<int> cost(n, -1);
  int m;
  cin >> m;
  for (int i = 0; i < m; ++i) {
    int a, b, c;
    cin >> a >> b >> c;
    if (cost[b - 1] == -1 || cost[b - 1] > c) {
      cost[b - 1] = c;
    }
  }
  int ans = 0, head = 0;
  for (int i = 0; i < n; ++i) {
    if (cost[i] != -1)
      ans += cost[i];
    else
      head++;
  }
  if (head > 1)
    cout << -1 << endl;
  else {
    if (!head) {
      int maxx = cost[0];
      for (int i = 1; i < n; ++i) maxx = max(maxx, cost[i]);
      ans -= maxx;
    }
    cout << ans << endl;
  }
  return 0;
}
