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:16777216")
using namespace std;
const int DIM = 1100;
int n, m;
int q[DIM];
vector<pair<int, int> > e[DIM];
int main() {
  scanf("%d", &n);
  for (int i = (0); i < (n); i++) {
    scanf("%d", &q[i]);
  }
  scanf("%d", &m);
  for (int i = (0); i < (m); i++) {
    int u, v, c;
    scanf("%d%d%d", &u, &v, &c);
    u--;
    v--;
    e[v].push_back(make_pair(c, u));
  }
  int ans = 0, cn = 0;
  for (int i = (0); i < (n); i++) {
    for (int j = (0); j < (n); j++) sort(e[j].begin(), e[j].end());
    if ((int)(e[i].size())) {
      ans += e[i][0].first;
      cn++;
    }
  }
  if (cn == n - 1)
    printf("%d\n", ans);
  else
    printf("-1\n");
  return 0;
}
