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 temp, i, j, k, T;
long long pow_r(long long d, long long n) {
  long long x = 1ll;
  while (n > 0) {
    if (n & 1) x = (x * d);
    d = (d * d);
    n >>= 1;
  }
  return x;
}
int expe[1005], e, mxexp, key[1005];
int n, m, a, b, c, cand;
long long ans;
bool done[1005];
int main() {
  scanf("%d", &n);
  for (i = 0; i < n; ++i) {
    key[i] = (int)1e9;
  }
  for (i = 0; i < n; ++i) scanf("%d", &expe[i]);
  scanf("%d", &m);
  for (i = 0; i < m; ++i) {
    scanf("%d", &a);
    scanf("%d", &b);
    scanf("%d", &c);
    --a;
    --b;
    key[b] = min(c, key[b]);
  }
  c = 0;
  for (i = 0; i < n; ++i) {
    if (key[i] == (int)1e9)
      ++c;
    else
      ans += key[i];
  }
  if (c > 1)
    puts("-1");
  else
    cout << ans << endl;
  return 0;
}
