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;
int val[10005];
int mark[10005];
bool vis[10005];
struct node {
  int a, b, c;
} a[10005];
bool cmp(node s1, node s2) { return s1.c < s2.c; }
int m;
int cnt = 0;
int ans = 0;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &val[i]);
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d %d %d", &a[i].a, &a[i].b, &a[i].c);
    if (val[a[i].a] <= val[a[i].b]) {
      i--;
      m--;
    }
  }
  sort(a + 1, a + 1 + m, cmp);
  for (int i = 1; i <= m; i++) {
    if (!vis[a[i].b]) {
      ans += a[i].c;
      vis[a[i].b] = 1;
    }
  }
  int sum = 0;
  for (int i = 1; i <= n; i++) {
    if (vis[i] == 0) {
      sum++;
    }
  }
  if (sum == 1) {
    printf("%d", ans);
  } else {
    printf("-1");
  }
  return 0;
}
