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 pr[1000000 + 7];
void init(int n) {
  for (int i = 0; i < n; i++) {
    pr[i] = i;
  }
}
int find(int i) {
  if (pr[i] == i) return i;
  return pr[i] = find(pr[i]);
}
bool Union(int i, int j) {
  int root1 = find(i);
  int root2 = find(j);
  if (root1 == root2) return false;
  pr[root1] = root2;
  return true;
}
struct Edge {
  int u, v;
  int w;
  bool operator<(const Edge &p) const { return w < p.w; }
};
bool visited[1000000 + 7] = {};
vector<Edge> e;
int sum = 0;
bool kruskal(int n) {
  int test = 0;
  sort(e.begin(), e.end());
  init(n);
  for (int i = 0; i < (int)e.size(); i++) {
    if (visited[e[i].v] == false) {
      if (Union(e[i].u, e[i].v)) {
        test++;
        sum += e[i].w;
        visited[e[i].v] = 1;
      }
    }
  }
  if (test < n - 1)
    return false;
  else
    return true;
}
int main() {
  int n;
  cin >> n;
  vector<int> q(n);
  for (int i = 0; i < n; i++) cin >> q[i];
  int m, a, b, c;
  cin >> m;
  for (int i = 0; i < m; i++) {
    cin >> a >> b >> c;
    Edge data;
    data.u = a;
    data.v = b;
    data.w = c;
    e.push_back(data);
  }
  if (kruskal(n) == false)
    cout << "-1" << endl;
  else {
    cout << sum << endl;
  }
}
