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;
struct Edge {
  int u, v, weight;
  bool operator<(const Edge& other) const { return weight < other.weight; }
};
int n;
int par[1005], qs[1005];
vector<pair<long long, int> > graph[1005];
int findRoot(int x) { return par[x] < 0 ? x : (par[x] = findRoot(par[x])); }
void merge(int a, int b) {
  a = findRoot(a);
  b = findRoot(b);
  if (a == b) return;
  if (par[b] < par[a]) swap(a, b);
  par[a] += par[b];
  par[b] = a;
}
bool sameSet(int a, int b) { return findRoot(a) == findRoot(b); }
int main() {
  int n, a, b, m, van = 0;
  long long cost = 0, c;
  bool pudo = false;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> qs[i];
  }
  cin >> m;
  priority_queue<pair<long long, int> > arcos;
  for (int i = 0; i < m; i++) {
    cin >> a >> b >> c;
    graph[a].push_back({-c, b});
    arcos.push({-c, b});
  }
  memset(par, -1, sizeof(par));
  van = 1;
  while (van < n && !arcos.empty()) {
    pair<long long, int> e = arcos.top();
    arcos.pop();
    if (!sameSet(0, e.second)) {
      cost += -e.first;
      van++;
      merge(0, e.second);
    }
  }
  if (van == n) {
    cout << cost << "\n";
  } else {
    cout << -1 << "\n";
  }
}
