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 MAXN = 1000 + 5;
const int INF = 1e9;
struct dude {
  int a, b, c;
} r[10 * MAXN];
int q[MAXN];
bool f(dude x, dude y) { return x.c < y.c; }
int main() {
  int n, m;
  scanf("%d", &n);
  vector<int> tree(n);
  for (int i = 0; i < n; ++i) tree[i] = i;
  for (int i = 0; i < n; ++i) scanf("%d", &q[i]);
  scanf("%d", &m);
  for (int i = 0; i < m; ++i) {
    int x, y;
    scanf("%d%d%d", &x, &y, &r[i].c);
    r[i].a = x - 1;
    r[i].b = y - 1;
  }
  sort(r, r + m, f);
  int cost = 0;
  int cnt = 0;
  for (int i = 0; i < m; ++i) {
    int x = r[i].a, y = r[i].b, val = r[i].c;
    if (tree[x] != tree[y] && tree[y] == y) {
      ++cnt;
      cost += val;
      int old_id = tree[y], new_id = tree[x];
      for (int j = 0; j < n; ++j)
        if (tree[j] == old_id) tree[j] = new_id;
    }
  }
  if (cnt == n - 1)
    printf("%d\n", cost);
  else
    printf("-1\n");
  return 0;
}
