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 n, root;
int sal[1111];
struct st {
  int a, b, c;
} s[11111];
vector<vector<int> > v;
vector<vector<int> > g;
vector<vector<int> > gc;
vector<vector<int> > cost;
void dfs(int vert, bool *f) {
  for (int i = 0; i < v[vert].size(); i++) {
    int u = v[vert][i];
    if (f[u] == 0) {
      f[u] = 1;
      dfs(u, f);
    }
  }
}
int main() {
  cin >> n;
  int ma = 0;
  v.clear();
  v.resize(n + 1);
  g.clear();
  g.resize(n + 1);
  gc.clear();
  gc.resize(n + 1);
  cost.clear();
  cost.resize(n + 1);
  for (int i = 0; i < n; i++) {
    cin >> sal[i];
    if (ma < sal[i]) {
      ma = sal[i];
      root = i + 1;
    }
  }
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    cin >> s[i].a >> s[i].b >> s[i].c;
    if (s[i].b == root) {
      cout << "-1" << endl;
      return 0;
    }
    v[s[i].a].push_back(s[i].b);
    cost[s[i].a].push_back(s[i].c);
    g[s[i].b].push_back(s[i].a);
    gc[s[i].b].push_back(s[i].c);
  }
  long long sum = 0;
  for (int i = 1; i <= n; i++)
    if (i != root) {
      ma = 100000000;
      for (int j = 0; j < g[i].size(); j++) ma = min(ma, gc[i][j]);
      if (ma == 100000000) {
        cout << "-1" << endl;
        return 0;
      }
      sum += ma;
    }
  cout << sum << endl;
}
