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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, c, e, x, y, root, cnt = 1;
  long long sum = 0;
  cin >> n;
  pair<int, int> p[n + 5];
  vector<int> v[n + 5];
  int a[n + 5], graph[n + 5][n + 5];
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      graph[i][j] = 998244353;
    }
  }
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    p[i] = make_pair(a[i], i + 1);
  }
  sort(p, p + n);
  cin >> e;
  for (int i = 0; i < e; i++) {
    cin >> x >> y >> c;
    v[y].push_back(x);
    graph[y][x] = min(c, graph[y][x]);
  }
  root = p[n - 1].second;
  for (int i = n - 2; i >= 0; i--) {
    int mn = 1000005, ind, f = 0;
    for (int j = 0; j < v[p[i].second].size(); j++) {
      int u = v[p[i].second][j];
      if (a[u - 1] > a[p[i].second - 1])
        mn = min(mn, graph[p[i].second][u]), f = 1;
    }
    if (f) cnt++, sum += mn;
  }
  if (cnt == n)
    cout << sum << endl;
  else
    cout << -1 << endl;
  return 0;
}
