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 MAXV = 1000;
struct edge {
  int s;
  int d;
  int w;
};
bool compare(edge *a, edge *b) { return a->w < b->w; }
int parent[MAXV];
int size[MAXV];
bool ok[MAXV];
int find_set(int a) {
  if (parent[a] == a) {
    return a;
  } else {
    parent[a] = find_set(parent[a]);
    return parent[a];
  }
}
void union_set(int a, int b) {
  if (size[a] < size[b]) {
    parent[a] = b;
    size[b] += size[a];
  } else {
    parent[b] = a;
    size[a] += size[b];
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i < n + 1; i++) {
    int q;
    cin >> q;
    parent[i] = i;
    size[i] = 1;
    ok[i] = true;
  }
  int m;
  cin >> m;
  vector<edge *> g;
  for (int i = 0; i < m; i++) {
    edge *e = new edge;
    cin >> e->s >> e->d >> e->w;
    g.push_back(e);
  }
  sort(g.begin(), g.end(), compare);
  long long int ans = 0;
  int cnt = 0;
  for (int i = 0; i < m; i++) {
    if (find_set(g[i]->s) != find_set(g[i]->d) && ok[g[i]->d]) {
      ok[g[i]->d] = false;
      union_set(find_set(g[i]->s), find_set(g[i]->d));
      ans += g[i]->w;
      cnt++;
    }
  }
  if (cnt == n - 1) {
    cout << ans << endl;
  } else {
    cout << -1 << endl;
  }
}
