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 main() {
  int n, a[1002], m, i, b[10001][3], max = 0, maxi = 1, c[1002] = {0}, d[1002],
                                     result = 0;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] > max) {
      max = a[i];
      maxi = i + 1;
    }
  }
  for (i = 0; i <= n; i++) {
    d[i] = 1000000;
  }
  cin >> m;
  for (i = 0; i < m; i++) {
    cin >> b[i][0] >> b[i][1] >> b[i][2];
    c[b[i][1]]++;
  }
  for (i = 1; i <= n; i++) {
    if (i != maxi) {
      if (c[i] == 0) {
        cout << "-1";
        return 0;
      }
    }
  }
  for (i = 0; i < m; i++) {
    if (b[i][2] < d[b[i][1]]) d[b[i][1]] = b[i][2];
  }
  for (i = 1; i <= n; i++) {
    if (i != maxi) result += d[i];
  }
  cout << result;
  return 0;
}
