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 mxn = 1024;
int q[mxn];
int p[mxn];
int haveParent[mxn];
int n;
int find(int x) {
  if (p[x] == x) return x;
  return p[x] = find(p[x]);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &q[i]);
  for (int i = 0; i < mxn; i++) p[i] = i;
  int m;
  scanf("%d", &m);
  vector<pair<int, pair<int, int> > > e;
  while (m--) {
    int a, b, c;
    scanf("%d", &a);
    scanf("%d", &b);
    scanf("%d", &c);
    e.push_back(make_pair(c, make_pair(a, b)));
  }
  sort(e.begin(), e.end());
  long long cost = 0;
  for (int i = 0; i < e.size(); i++) {
    int C = e[i].first;
    int A = e[i].second.first;
    int B = e[i].second.second;
    if (find(A) == find(B)) continue;
    if (haveParent[B]) continue;
    haveParent[B] = 1;
    p[B] = A;
    cost += C;
  }
  bool ok = 1;
  for (int i = 2; i <= n; i++)
    if (find(i) != find(1)) {
      ok = 0;
      break;
    }
  if (!ok) cost = -1;
  cout << cost << endl;
  return 0;
}
