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;
vector<int> setunion;
int findHead(int a) {
  if (setunion[a] == a) return a;
  return setunion[a] = findHead(setunion[a]);
}
int main() {
  long long n;
  cin >> n;
  vector<long long> qual(n);
  for (long long i{}; i < n; ++i) cin >> qual[i];
  long long m;
  cin >> m;
  priority_queue<array<int, 3>, vector<array<int, 3>>, greater<array<int, 3>>>
      pq;
  setunion = vector<int>(n);
  for (int i{}; i < n; ++i) setunion[i] = i;
  for (long long i{}; i < m; ++i) {
    int a, b, c;
    cin >> a >> b >> c;
    pq.push({c, a - 1, b - 1});
  }
  long long out{};
  map<int, int> dist;
  while (!pq.empty()) {
    auto v = pq.top();
    pq.pop();
    if (findHead(v[1]) != findHead(v[2])) {
      if (setunion[v[2]] != v[2]) {
        if (v[0] < dist[v[2]]) {
          out -= dist[v[2]];
          out += v[0];
          setunion[v[2]] = setunion[v[1]];
        }
        continue;
      }
      out += v[0];
      setunion[v[2]] = setunion[v[1]];
    }
  }
  for (int i{}; i < n - 1; ++i) {
    if (findHead(i) != findHead(i + 1)) {
      cout << -1 << '\n';
      return 0;
    }
  }
  cout << out << '\n';
}
