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;
struct info {
  int u, v, x;
  info() {}
  info(int uu, int vv, int xx) {
    u = uu;
    v = vv;
    x = xx;
  }
  bool operator<(const info &b) const { return x < b.x; }
};
int n, m, x, sz[1010], link[1010];
info app[10000];
bool ase[1010];
int findLink(int x) {
  while (x != link[x]) x = link[x];
  return x;
}
bool isSameLink(int a, int b) { return findLink(a) == findLink(b); }
void unite(int a, int b) {
  a = findLink(a);
  b = findLink(b);
  if (a == b) return;
  if (sz[a] < sz[b]) swap(a, b);
  sz[a] += sz[b];
  link[b] = a;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) sz[i] = 1, link[i] = i;
  for (int i = 0; i < n; i++) scanf("%d", &x);
  scanf("%d", &m);
  for (int i = 0; i < m; i++)
    scanf("%d %d %d", &app[i].u, &app[i].v, &app[i].x);
  sort(app, app + m);
  int ans = 0;
  for (int i = 0; i < m; i++) {
    int u = app[i].u, v = app[i].v, w = app[i].x;
    if (ase[v] || isSameLink(u, v)) continue;
    ans += w;
    unite(u, v);
    ase[v] = 1;
  }
  for (int i = 1; i < n; i++) {
    if (!isSameLink(i, i + 1)) {
      cout << -1 << endl;
      return 0;
    }
  }
  cout << ans << endl;
}
