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;
struct edge {
  int a, b;
  int w;
  bool operator<(edge const& e) const { return w < e.w; }
};
int n, m;
int p[1500];
bool has[1500];
edge e[10 * 1500];
int ans;
int get(int a) { return (a == p[a] ? a : p[a] = get(p[a])); }
void unite(int a, int b) { p[get(a)] = p[get(b)]; }
int main() {
  cin >> n;
  for (int q, i = 0; i < n; ++i) {
    cin >> q;
    p[i] = i;
  }
  cin >> m;
  for (int i = 0; i < m; ++i) {
    cin >> e[i].a >> e[i].b >> e[i].w;
    --e[i].a;
    --e[i].b;
  }
  sort(e, e + m);
  for (int i = 0; i < m; ++i) {
    if (!has[e[i].b]) {
      if (get(e[i].a) != get(e[i].b)) {
        unite(e[i].a, e[i].b);
        ans += e[i].w;
        has[e[i].b] = true;
      }
    }
  }
  int cnt = 0;
  for (int i = 0; i < n; ++i) {
    cnt += (i == p[i]);
  }
  cout << (cnt > 1 ? -1 : ans);
  return 0;
}
