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>
#pragma comment(linker, "/STACK:16000000")
using namespace std;
const int maxn = 30010;
int col[maxn] = {0};
int inp[maxn] = {0};
struct pnt {
  int from;
  int to;
  int cost;
};
pnt mas[maxn];
int q[maxn] = {0};
bool cmp(pnt a, pnt b) { return a.cost < b.cost; }
int main() {
  int i, j, n, m, tmp;
  cin >> n;
  for (i = 1; i <= n; i++) cin >> q[i];
  cin >> m;
  int a, b, c;
  for (i = 1; i <= m; i++) {
    cin >> mas[i].from >> mas[i].to >> mas[i].cost;
  }
  if (n == 1) {
    cout << "0";
    return 0;
  }
  sort(mas + 1, mas + 1 + m, cmp);
  long long res = 0;
  for (i = 1; i <= n; i++) col[i] = i;
  for (i = 1; i <= m; i++) {
    a = mas[i].from;
    b = mas[i].to;
    if (col[a] != col[b] && inp[b] == 0) {
      res += (long long)mas[i].cost;
      inp[b] = 1;
      tmp = col[a];
      for (j = 1; j <= n; j++)
        if (col[j] == tmp) col[j] = col[b];
    }
  }
  int cnt = 0;
  for (i = 1; i <= n; i++)
    if (inp[i] == 0) {
      cnt++;
    }
  if (cnt == 1)
    cout << res;
  else
    cout << "-1";
  return 0;
}
