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;
int n, m, a, b, c, x, res, d[1001], k[1001], inl[1001];
vector<int> edg[1001], cost[1001], q;
void dfs(int x) {
  k[x] = 1;
  for (int i = 0; i < (edg[x].size()); i++) {
    d[edg[x][i]] = min(d[edg[x][i]], cost[x][i]);
    if (!k[edg[x][i]]) dfs(edg[x][i]);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < (n); i++) {
    d[i + 1] = (1 << 30);
    scanf("%d", &a);
    q.push_back(a);
  }
  scanf("%d", &m);
  for (int i = 0; i < (m); i++) {
    scanf("%d%d%d", &a, &b, &c);
    if (q[a - 1] > q[b - 1]) {
      edg[a].push_back(b);
      cost[a].push_back(c);
      inl[b]++;
    }
  }
  for (int i = (1); i <= (n); i++) {
    if (inl[i] == 0) {
      d[i] = 0;
      dfs(i);
      x++;
    }
  }
  for (int i = (1); i <= (n); i++) res += d[i];
  if (x != 1)
    printf("-1\n");
  else
    printf("%d\n", res);
  return 0;
}
