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;
long long rar[10100], s[10100], a[10100], sp[10100];
long long root(long long x) {
  while (rar[x] != x) {
    rar[x] = rar[rar[x]];
    x = rar[x];
  }
  return x;
}
void un(long long a, long long b) {
  long long rta = root(a);
  long long rtb = root(b);
  if (s[rta] > s[rtb]) {
    rar[rtb] = rta;
    s[rta] += s[rtb];
  } else {
    rar[rta] = rtb;
    s[rtb] += s[rta];
  }
}
int main() {
  for (long long i = 1; i <= 1010; i++) {
    rar[i] = i;
    s[i] = 1;
  }
  long long n, m, x, b, c;
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  cin >> m;
  pair<long long, pair<long long, long long> > p[m + 1];
  for (long long i = 1; i <= m; i++) {
    cin >> x >> b >> c;
    p[i - 1] = make_pair(c, make_pair(x, b));
  }
  sort(p, p + m);
  long long sm = 0;
  for (long long i = 0; i <= m; i++) {
    c = p[i].first;
    x = p[i].second.first;
    b = p[i].second.second;
    if (root(x) != root(b) && !sp[b]) {
      sp[b] = 1;
      un(x, b);
      sm += c;
    }
  }
  if (s[root(1)] != n)
    cout << -1 << endl;
  else
    cout << sm << endl;
}
