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 oo = 1 << 30;
bool cmp1(pair<int, vector<pair<int, int> > > a,
          pair<int, vector<pair<int, int> > > b) {
  return a.first < b.first;
}
bool cmp2(pair<int, int> a, pair<int, int> b) { return a.first < b.first; }
int main() {
  int n;
  scanf("%d", &n);
  vector<pair<int, vector<pair<int, int> > > > Q(n);
  for (int i = 0; i < n; i++) scanf("%d", &Q[i].first);
  int m, ai, bi, ci;
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d", &ai, &bi, &ci);
    ai--;
    bi--;
    if (Q[ai].first > Q[bi].first) Q[bi].second.push_back(make_pair(ci, ai));
  }
  sort(Q.rbegin(), Q.rend(), cmp1);
  for (int i = 0; i < n; i++)
    sort(Q[i].second.begin(), Q[i].second.end(), cmp2);
  if (Q[0].second.size() != 0)
    printf("-1\n");
  else {
    int minc = 0;
    for (int i = 1; i < n; i++)
      if (Q[i].second.size() == 0) {
        minc = -1;
        break;
      } else
        minc += Q[i].second[0].first;
    printf("%d\n", minc);
  }
}
