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 q[1010];
vector<int> p[1010];
vector<int> c[1010];
bool cmp(int a, int b) { return q[a] > q[b]; }
int a[1010];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &q[i]);
  int m;
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    int a, b, w;
    scanf("%d %d %d", &a, &b, &w);
    --a, --b;
    p[b].push_back(a);
    c[b].push_back(w);
  }
  for (int i = 0; i < n; i++) a[i] = i;
  sort(a, a + n, cmp);
  int ans = 0;
  bool q = false;
  for (int i = 0; i < n; i++) {
    int v = a[i];
    int sz = p[v].size();
    int tans = 1 << 25;
    for (int i = 0; i < sz; i++) tans = min(tans, c[v][i]);
    if (tans == (1 << 25)) {
      if (q) {
        printf("-1\n");
        return 0;
      }
      q = true;
      continue;
    }
    ans += tans;
  }
  cout << ans << endl;
  return 0;
}
