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;
ifstream f("date.in");
vector<pair<int, int> > v[5000];
vector<pair<int, pair<int, int> > > muc;
bitset<5000> viz;
queue<int> q;
int n, m, a[5000], T[5000], total, gr[5000];
void read() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int x, y, c;
    cin >> x >> y >> c;
    v[x].push_back({c, y});
    muc.push_back({c, {x, y}});
  }
}
int getDad(int x) {
  if (T[x] == x)
    return x;
  else
    return T[x] = getDad(T[x]);
}
int c;
int main() {
  read();
  if (m == 0) {
    if (n == 1)
      cout << 0;
    else
      cout << -1;
    return 0;
  }
  for (int i = 1; i <= n; i++) T[i] = i;
  sort(muc.begin(), muc.end());
  for (int i = 0; i < muc.size(); i++) {
    int n1 = muc[i].second.first;
    int n2 = muc[i].second.second;
    int cost = muc[i].first;
    int tn1 = getDad(n1);
    int tn2 = getDad(n2);
    if (a[n1] > a[n2] && tn1 != tn2 && gr[n2] == 0) {
      T[n1] = tn2;
      total += cost;
      viz[n1] = viz[n2] = 1;
      gr[n2]++;
      c++;
    }
  }
  if (c != n - 1) {
    cout << -1;
    return 0;
  }
  for (int i = 1; i <= n; i++)
    if (viz[i] == 0) {
      cout << -1;
      return 0;
    }
  cout << total;
}
