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;
long long par[2000];
long long sz[2000];
void make_set() {
  for (long long i = 0; i < 2000; i++) {
    par[i] = i;
    sz[i] = 1;
  }
}
long long find_set(long long a) {
  if (par[a] == a) return a;
  return find_set(par[a]);
}
void union_set(long long a, long long b) {
  a = find_set(a);
  b = find_set(b);
  par[b] = a;
  sz[a] += sz[b];
}
int main() {
  long long n;
  cin >> n;
  make_set();
  long long a[n];
  for (long long i = 0; i < n; i++) cin >> a[i];
  long long m;
  cin >> m;
  vector<vector<pair<long long, long long>>> adj(n);
  for (long long i = 0; i < m; i++) {
    long long x, y, w;
    cin >> x >> y >> w;
    x--;
    y--;
    adj[x].push_back(make_pair(y, w));
  }
  priority_queue<pair<long long, pair<long long, long long>>,
                 vector<pair<long long, pair<long long, long long>>>,
                 greater<pair<long long, pair<long long, long long>>>>
      q;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < adj[i].size(); j++) {
      q.push(make_pair(adj[i][j].second, make_pair(i, adj[i][j].first)));
    }
  }
  long long cost = 0;
  while (!q.empty()) {
    pair<long long, pair<long long, long long>> p = q.top();
    q.pop();
    long long w = p.first;
    long long x = p.second.first;
    long long y = p.second.second;
    if (par[y] == y && find_set(x) != find_set(y)) {
      union_set(x, y);
      cost += w;
    }
  }
  long long cnt = 0;
  for (long long i = 0; i < n; i++) {
    if (par[i] == i) cnt++;
  }
  if (cnt > 1) {
    cout << -1 << endl;
    return 0;
  }
  cout << cost << endl;
}
