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;
void ga(int N, int *A) {
  for (int i(0); i < N; i++) scanf("%d", A + i);
}
struct nd {
  int a, b, w, is;
  bool operator<(const nd &r) const { return w < r.w; }
} g[10005];
int C[10005];
int spa(int M, int N) {
  int S = 0, W = N - 1;
  sort(g, g + M);
  for (int i(0); i < M && W; i++)
    if (!C[g[i].b]) S += g[i].w, C[g[i].b] = 1, --W;
  return W ? -1 : S;
}
int N, M, A[10005], a, b, c;
int main(void) {
  scanf("%d", &N), ga(N, A);
  scanf("%d", &M);
  for (int i(0); i < M; i++)
    scanf("%d%d%d", &a, &b, &c), g[i] = {--a, --b, c, 0};
  printf("%d\n", spa(M, N));
  return 0;
}
