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;
using namespace std;
void chekarre(int* arr, int n) {
  cout << "[";
  for (int i = 0; i < n; i++) cout << arr[i] << " ";
  cout << "]" << endl;
}
int n, m;
int can[1003] = {0};
int temp[1004] = {0};
vector<pair<int, pair<int, int> > > pool;
int solve() {
  int a, b, c, sum = 0;
  if (n == 1) {
    cout << "0";
    return 1;
  }
  sort(pool.begin(), pool.end());
  int cnt = 0;
  for (int i = 0; i < pool.size(); i++) {
    a = pool[i].second.first;
    b = pool[i].second.second;
    c = pool[i].first;
    if (can[b] == 0) {
      sum += c, can[b] = a;
      cnt++;
    }
  }
  if (cnt != n - 1) {
    cout << -1;
    return 1;
  }
  cout << sum;
  return 1;
}
bool input() {
  scanf("%d", &n);
  int a;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a);
  }
  scanf("%d", &m);
  int b, c;
  for (int i = 0; i < m; i++) {
    scanf("%d", &a);
    scanf("%d", &b);
    scanf("%d", &c);
    pool.push_back(pair<int, pair<int, int> >(c, pair<int, int>(a, b)));
  }
  return true;
}
int main() {
  int T = 1;
  for (int testnum = 1; testnum <= T; testnum++) {
    if (!input()) break;
    solve();
  }
  return 0;
}
