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 int parent[1001];
long long int find_parent(long long int x) {
  if (x == parent[x]) return x;
  return find_parent(parent[x]);
}
bool unite(long long int x, long long int y) {
  long long int a = parent[x], b = parent[y];
  if (b != y) return false;
  parent[y] = x;
  return true;
}
void solve() {
  long long int n, i, j;
  cin >> n;
  long long int q[n];
  for (i = 0; i < n; i++) cin >> q[i];
  long long int m;
  cin >> m;
  vector<pair<long long int, pair<long long int, long long int>>> v;
  for (i = 0; i < m; i++) {
    long long int x, y, c;
    cin >> x >> y >> c;
    v.push_back({c, {x, y}});
  }
  for (long long int i = 1; i <= n; i++) parent[i] = i;
  sort(v.begin(), v.end());
  long long int ans = 0;
  for (long long int i = 0; i < m; i++) {
    long long int c = v[i].first, x = v[i].second.first, y = v[i].second.second;
    long long int a = find_parent(x), b = find_parent(y);
    if (a != b && unite(x, y)) {
      ans += c;
    }
  }
  for (long long int i = 2; i <= n; i++) {
    if (find_parent(i) != find_parent(1)) {
      cout << "-1";
      return;
    }
  }
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  t = 1;
  while (t--) solve();
  return 0;
}
