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, counter;
int q[1005];
int mtx[1005][1005];
int arr[3];
int ans = 0;
bool vis[1001];
int main() {
  memset(mtx, -1, sizeof(mtx));
  memset(vis, 0, sizeof(vis));
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &q[i]);
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf("%d %d %d", &arr[0], &arr[1], &arr[2]);
    if ((arr[2] < mtx[arr[0] - 1][arr[1] - 1] ||
         mtx[arr[0] - 1][arr[1] - 1] == -1) &&
        q[arr[0] - 1] > q[arr[1] - 1]) {
      vis[arr[1] - 1] = true;
      mtx[arr[0] - 1][arr[1] - 1] = arr[2];
    }
  }
  for (int i = 0; i < n; i++) {
    int mini = 0x3f3f3f;
    bool check = false;
    for (int j = 0; j < n; j++) {
      if (mtx[j][i] != -1) {
        check = true;
        mini = min(mtx[j][i], mini);
      }
    }
    if (!check) mini = 0;
    ans += mini;
  }
  for (int i = 0; i < n; i++) {
    if (!vis[i]) ++counter;
  }
  if (counter > 1) return !printf("-1\n");
  printf("%d\n", ans);
  return 0;
}
