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() {
  int n;
  cin >> n;
  vector<int> q(n);
  for (int i = 0; i < n; i++) cin >> q[i];
  int m = 0;
  cin >> m;
  vector<vector<pair<int, int> > > reb(n);
  vector<vector<pair<int, int> > > reb_rev(n);
  for (int i = 0; i < m; i++) {
    int j = 0;
    cin >> j;
    int b, c;
    cin >> b >> c;
    reb[j - 1].push_back(make_pair(b - 1, c));
    reb_rev[b - 1].push_back(make_pair(j - 1, c));
  }
  vector<bool> is(n);
  int k = 0;
  int res = 0;
  while (k != n - 1) {
    int cur = 0;
    for (int i = 0; i < n; i++) {
      if (reb[i].empty()) {
        if (is[i]) continue;
        cur = i;
        break;
      }
    }
    k++;
    is[cur] = 1;
    if (reb_rev[cur].empty()) {
      cout << -1 << endl;
      return 0;
    }
    int mn = reb_rev[cur][0].second;
    for (int i = 0; i < reb_rev[cur].size(); i++)
      if (mn > reb_rev[cur][i].second) mn = reb_rev[cur][i].second;
    res += mn;
    for (int i = 0; i < reb_rev[cur].size(); i++) {
      for (int j = 0; j < reb[reb_rev[cur][i].first].size(); j++) {
        if (reb[reb_rev[cur][i].first][j].first == cur) {
          reb[reb_rev[cur][i].first].erase(reb[reb_rev[cur][i].first].begin() +
                                           j);
          j--;
        }
      }
    }
    reb_rev[cur].clear();
  }
  cout << res << endl;
}
