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;
vector<long long> papa;
vector<long long> cost;
void init(long long n) {
  for (int i = 0; i < n; i++) {
    papa.push_back(i);
    cost.push_back(INT_MAX);
  }
}
long long Find(long long j) {
  while (papa[j] != j) {
    j = papa[j];
  }
  return j;
}
void Union(long long b, long long a, long long c) {
  papa[a] = Find(b);
  cost[a] = min(c, cost[a]);
}
int main(int argc, char const *argv[]) {
  long long n;
  cin >> n;
  init(n);
  vector<long long> q(n);
  for (int i = 0; i < n; i++) {
    cin >> q[i];
  }
  long long m;
  cin >> m;
  long long x, y, c;
  while (m--) {
    cin >> x >> y >> c;
    x--, y--;
    Union(x, y, c);
  }
  long long cnt = 0;
  for (int i = 0; i < n; i++) {
    if (papa[i] == i) cnt++;
  }
  if (cnt != 1) {
    cout << -1;
  } else {
    long long co = INT_MAX * -1;
    for (int i = 0; i < n; i++) {
      co += cost[i];
    }
    cout << co;
  }
  return 0;
}
