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;
vector<pair<int, int>> v[100001];
int visited[100001] = {0};
int ct = 0;
long long int power(long long int a, long long int b, long long int mod) {
  a %= mod;
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return res;
}
int sum = 0;
int dfs(int x, int a[]) {
  for (auto y : v[x]) {
    if (a[x] < a[y.second]) {
      visited[x] = 1;
      sum += y.first;
      return 1;
    }
  }
  return 0;
}
int main() {
  int n, i, x, y;
  cin >> n;
  int a[n];
  for (i = 0; i < n; i++) cin >> a[i];
  int q;
  cin >> q;
  set<int> s;
  while (q--) {
    int ii;
    cin >> x >> y >> ii;
    x -= 1;
    y -= 1;
    s.insert(y);
    v[y].push_back({ii, x});
  }
  for (auto x : s) {
    sort(v[x].begin(), v[x].end());
    if (!dfs(x, a)) {
      cout << -1;
      return 0;
    }
  }
  int cnt = 0;
  for (i = 0; i < n; i++) {
    if (visited[i] == 0) {
      cnt += 1;
    }
    if (cnt > 1) {
      cout << -1;
      return 0;
    }
  }
  if (cnt == 1)
    cout << sum;
  else
    cout << -1;
}
