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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  std::cerr << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  std::cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
using graph = vector<vector<int>>;
ostream& operator<<(ostream& os, const pair<int, int>& p) {
  os << "(" << p.first << ", " << p.second << ")";
  return os;
}
ostream& operator<<(ostream& os, const vector<int>& p) {
  for (auto a : p) os << a << ' ';
  return os;
}
ostream& operator<<(ostream& os, const graph& g) {
  for (int i = 0; i < g.size(); ++i, os << endl) {
    os << i << ": " << g[i];
  }
  return os;
}
const long long MOD = 1e+9 + 7;
const long long INFLL = 0x3f3f3f3f3f3f3f3f;
const int INF = 0x3f3f3f3f;
const long long MAXN = 1e+5 + 7;
void solve() {
  int n, m;
  cin >> n;
  int has[n];
  for (int i = 0, c; i < n; ++i) cin >> c;
  memset(has, 0, sizeof has);
  cin >> m;
  long long ans = 0;
  vector<pair<int, pair<int, int>>> v(m);
  for (int i = 0; i < m; ++i) {
    int a, b, c;
    cin >> a >> b >> c;
    a--, b--;
    v[i] = {c, {a, b}};
  }
  sort(begin(v), end(v));
  for (int i = 0; i < m; ++i) {
    if (!has[v[i].second.second]) {
      ans += v[i].first;
      has[v[i].second.second] = 1;
    }
  }
  int cnt = 0;
  for (int i = 0; i < n; ++i) {
    cnt += !has[i];
  }
  cout << (cnt > 1 ? -1 : ans) << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long i, j, n;
  int t = 1;
  while (t--) solve();
  return 0;
}
