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;
string lower(string s) {
  for (int i = 0; i < s.size(); i++) s[i] = tolower(s[i]);
  return s;
}
vector<string> sep(string s, char c) {
  string temp;
  vector<string> res;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == c) {
      if (temp != "") res.push_back(temp);
      temp = "";
      continue;
    }
    temp = temp + s[i];
  }
  if (temp != "") res.push_back(temp);
  return res;
}
template <class T>
T toint(string s) {
  stringstream ss(s);
  T ret;
  ss >> ret;
  return ret;
}
template <class T>
string tostr(T inp) {
  string ret;
  stringstream ss;
  ss << inp;
  ss >> ret;
  return ret;
}
template <class T>
void D(T A[], int n) {
  for (int i = 0; i < n; i++) cout << A[i] << " ";
  cout << endl;
}
template <class T>
void D(vector<T> A, int n = -1) {
  if (n == -1) n = A.size();
  for (int i = 0; i < n; i++) cout << A[i] << " ";
  cout << endl;
}
vector<int> edge[1001];
vector<int> cost[1001];
int main() {
  int n;
  scanf("%d", &n);
  vector<pair<int, int> > lst;
  for (int(i) = (0); (i) < (n); (i)++) {
    int pq;
    scanf("%d", &pq);
    lst.push_back((make_pair(pq, i)));
  }
  sort(lst.rbegin(), lst.rend());
  int m;
  scanf("%d", &m);
  for (int(i) = (0); (i) < (m); (i)++) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    a--;
    b--;
    edge[b].push_back((a));
    cost[b].push_back((c));
  }
  long long ans = 1000000000000LL;
  for (int(i) = (0); (i) < (n); (i)++) {
    long long total = 0;
    bool tval = 1;
    for (int(j) = (0); (j) < (n); (j)++) {
      if (i == j) continue;
      if (edge[j].size() == 0) {
        tval = 0;
        break;
      }
      long long chs = 1000000000000LL;
      for (int(k) = (0); (k) < (edge[j].size()); (k)++)
        chs = min(chs, (long long)cost[j][k]);
      total += chs;
    }
    if (!tval) continue;
    ans = min(ans, total);
  }
  if (ans == 1000000000000LL)
    cout << -1 << endl;
  else
    cout << ans << endl;
}
