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;
const long double eps = 1e-9;
vector<int> w;
vector<bool> passed;
vector<vector<pair<int, int> > > g, ig;
int n;
void dfs(int u) {
  passed[u] = true;
  for (int i = 0; i < g[u].size(); ++i)
    if (!passed[g[u][i].first] && w[g[u][i].first] == u) dfs(g[u][i].first);
}
int main() {
  cin >> n;
  vector<int> q(n);
  for (int i = 0; i < n; ++i) cin >> q[i];
  int m;
  scanf("%d", &m);
  vector<int> d(n, 1000000000);
  w.resize(n);
  passed.resize(n);
  g.resize(n);
  ig.resize(n);
  for (int i = 0; i < m; ++i) {
    int u, v, c;
    scanf("%d%d%d", &u, &v, &c);
    u--;
    v--;
    g[u].push_back(make_pair(v, c));
    ig[v].push_back(make_pair(u, c));
  }
  int root = max_element(q.begin(), q.end()) - q.begin();
  int cost = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < ig[i].size(); ++j) {
      if (d[i] > ig[i][j].second) {
        d[i] = ig[i][j].second;
        w[i] = ig[i][j].first;
      }
    }
    if (i != root) cost += d[i];
  }
  dfs(root);
  int cnt = 0;
  for (int i = 0; i < n; ++i) cnt += passed[i];
  if (cnt != n)
    cout << -1 << endl;
  else
    cout << cost << endl;
  return 0;
}
