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;
class dsu {
 public:
  vector<long long> parent;
  vector<long long> size;
  void init(long long a) {
    parent.resize(a);
    size.resize(a);
    for (long long i = 0; i < a; i++) {
      parent[i] = i;
      size[i] = 1;
    }
  }
  long long par(long long i) {
    if (i == parent[i]) return i;
    return parent[i] = par(parent[i]);
  }
  void unite(long long a, long long b) {
    a = par(a);
    b = par(b);
    if (a != b) {
      parent[a] = b;
      size[b] += size[a];
      size[a] = 0;
    }
  }
};
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  vector<long long> v(n);
  for (long long i = 0; i < n; i++) cin >> v[i];
  vector<pair<long long, pair<long long, long long>>> pq;
  long long q;
  cin >> q;
  for (long long i = 0; i < q; i++) {
    long long a, b, w;
    cin >> a >> b >> w;
    pq.push_back({w, {a, b}});
  }
  dsu g;
  g.init(n);
  sort(pq.begin(), pq.end());
  long long ans = 0;
  vector<long long> visited(n + 1, 0);
  for (auto c : pq) {
    if (visited[c.second.second] != 1) {
      if (g.par(c.second.first - 1) != g.par(c.second.second - 1)) {
        ans += c.first;
        g.unite(c.second.first - 1, c.second.second - 1);
        visited[c.second.second] = 1;
      }
    }
  }
  long long cnt = 0;
  for (auto c : g.size) {
    if (c != 0) cnt++;
  }
  if (cnt == 1) {
    cout << ans << "\n";
  } else {
    cout << -1 << "\n";
  }
  return 0;
}
