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 int N = 1e3 + 200;
int n, m, u, v, c, cnt;
vector<pair<int, int> > g[N];
unordered_map<int, bool> visited;
int q[N];
const int fx[9] = {0, 0, 1, -1, 0, 1, 1, -1, -1};
const int fy[9] = {1, -1, 0, 0, 0, 1, -1, 1, -1};
const int dx[] = {0, 1, 0, -1};
const int dy[] = {1, 0, -1, 0};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> q[i];
  }
  cin >> m;
  for (int i = 1; i <= m; i++) {
    cin >> u >> v >> c;
    g[v].push_back({u, c});
  }
  int mn = INT_MAX;
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    for (auto j : g[i]) {
      mn = min(mn, j.second);
    }
    if (mn == INT_MAX) {
      cnt++;
      mn = 0;
    }
    ans += mn;
    mn = INT_MAX;
  }
  if (cnt > 1) {
    cout << -1;
    return 0;
  }
  cout << ans;
}
