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;
int parent[1010];
int root(int x) {
  if (x == parent[x])
    return x;
  else
    return parent[x] = root(parent[x]);
}
int main() {
  int n;
  cin >> n;
  bool visited[n];
  int qul[n];
  for (int i = 0; i < n; i++) {
    parent[i] = i;
    visited[i] = false;
    cin >> qul[i];
  }
  int m;
  int w, a, b;
  int t = 0;
  cin >> m;
  vector<pair<int, pair<int, int> > > edgs;
  for (int i = 0; i < m; i++) {
    cin >> a >> b >> w;
    a--;
    b--;
    edgs.push_back(make_pair(w, make_pair(a, b)));
  }
  sort(edgs.begin(), edgs.end());
  for (int i = 0; i < m; i++) {
    w = edgs[i].first;
    a = edgs[i].second.first;
    b = edgs[i].second.second;
    if (root(a) != root(b) && !visited[b]) {
      parent[b] = root(a);
      t += w;
      visited[b] = true;
    }
  }
  bool res = true;
  int par = root(0);
  for (int i = 0; i < n; i++) {
    if (root(i) != par) res = false;
  }
  if (res)
    cout << t;
  else
    cout << -1;
  return 0;
}
