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 n, m;
struct pp {
  int index, va;
} p[1000];
int MMM[1000][1000];
bool cmp(pp a, pp b) { return a.va > b.va; }
int minv[1000];
bool used[1000];
int main() {
  int a, b, c;
  memset(MMM, -1, sizeof(MMM));
  memset(minv, -1, sizeof(minv));
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> p[i].va;
    p[i].index = i;
  }
  sort(p, p + n, cmp);
  cin >> m;
  for (int i = 0; i < m; i++) {
    cin >> a >> b >> c;
    a--, b--;
    if (MMM[a][b] == -1 || c < MMM[a][b]) MMM[a][b] = c;
  }
  minv[p[0].index] = 0;
  used[p[0].index] = true;
  int ans = 0;
  int i;
  for (i = 0; i < n; i++) {
    if (!used[p[i].index]) {
      cout << "-1" << endl;
      break;
    }
    for (int j = i + 1; j < n; j++)
      if (MMM[p[i].index][p[j].index] >= 0) {
        used[p[j].index] = true;
        if (minv[p[j].index] == -1 ||
            minv[p[j].index] > MMM[p[i].index][p[j].index])
          minv[p[j].index] = MMM[p[i].index][p[j].index];
      }
  }
  if (i == n) {
    for (int j = 0; j < n; j++) ans += minv[j];
    cout << ans << endl;
  }
  return 0;
}
