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;
vector<pair<int, pair<int, int> > > e;
int Par[1010];
int find(int x) { return Par[x] = (Par[x] == x) ? x : find(Par[x]); }
void init() {
  for (int i = 0; i < 1010; i++) Par[i] = i;
}
int n, m;
int ans, cnt;
bool in[1010];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int t;
    scanf("%d", &t);
  }
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    e.push_back(make_pair(w, make_pair(u, v)));
  }
  sort(e.begin(), e.end());
  init();
  int len = e.size();
  for (int i = 0; i < len; i++) {
    if (in[e[i].second.second]) continue;
    int t1, t2;
    t1 = find(e[i].second.first);
    t2 = find(e[i].second.second);
    if (t1 == t2) continue;
    ans += e[i].first;
    cnt++;
    in[e[i].second.second] = 1;
    Par[t1] = t2;
  }
  if (cnt != n - 1)
    puts("-1");
  else
    printf("%d\n", ans);
  return 0;
}
