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>
template <class T>
using pt = std::pair<T, T>;
using namespace std;
vector<vector<bool>> g;
struct hoba {
  int s, f, c;
};
vector<hoba> ed;
struct snmchka {
  int g;
  snmchka() {
    static int ind = 0;
    g = ind++;
  }
};
vector<snmchka> snm;
int findik(int x) {
  if (snm[x].g != x) {
    snm[x].g = findik(snm[x].g);
  }
  return snm[x].g;
}
void unionchik(int x, int y) {
  x = findik(x);
  snm[y].g = x;
  return;
}
bool good(int x) { return findik(x) == x; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  g.resize(n);
  int i1, i2, c;
  int m;
  for (int i = 0; i < n; i++) {
    cin >> i1;
  }
  cin >> m;
  ed.resize(m);
  for (int i = 0; i < m; i++) {
    cin >> ed[i].s >> ed[i].f >> ed[i].c;
    ed[i].s--;
    ed[i].f--;
  }
  sort(ed.begin(), ed.end(), [](hoba a, hoba b) { return a.c < b.c; });
  long long sum = 0;
  snm.resize(n);
  for (int i = 0; i < m; i++) {
    if (good(ed[i].f)) {
      unionchik(ed[i].s, ed[i].f);
      sum += ed[i].c;
    }
  }
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    if (findik(i) == i) cnt++;
  }
  if (cnt > 1)
    cout << -1;
  else
    cout << sum << '\n';
  return 0;
}
