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;
pair<long long, pair<long long, long long> > edge[100001];
long long parent[1001];
long long visited[1001];
long long find_set(long long node) {
  if (parent[node] == node)
    return node;
  else
    return find_set(parent[node]);
}
void union_set(long long src, long long des) { parent[src] = parent[des]; }
int main() {
  long long n, m, ans = 0;
  cin >> n;
  long long q[n + 1];
  for (long long i = 1; i <= n; i++) {
    cin >> q[i];
    parent[i] = i;
  }
  cin >> m;
  for (long long i = 0; i < m; i++) {
    long long x, y, c;
    cin >> x >> y >> c;
    edge[i] = make_pair(c, make_pair(x, y));
  }
  sort(edge, edge + m);
  for (long long i = 0; i < m; i++) {
    long long src, des, val;
    src = edge[i].second.first;
    des = edge[i].second.second;
    val = edge[i].first;
    long long x = find_set(src);
    long long y = find_set(des);
    if (x != y && visited[des] == 0) {
      ans += val;
      union_set(x, y);
      visited[des] = 1;
    }
  }
  for (long long i = 1; i <= n; i++) {
    if (find_set(i) != find_set(1)) {
      cout << "-1" << endl;
      return 0;
    }
  }
  cout << ans << endl;
}
