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[1005], sz[1005], arr[1005];
bool vis[1005];
long long find(long long x) {
  if (par[x] == x) {
    return x;
  }
  return par[x] = find(par[x]);
}
void union_(long long x, long long y) {
  long long a = find(x);
  long long b = find(y);
  if (a == b) {
    return;
  }
  if (sz[a] > sz[b]) {
    par[b] = a;
    sz[a] += sz[b];
  } else {
    par[a] = b;
    sz[b] += sz[a];
  }
}
bool compare(pair<long long, pair<long long, long long>> a,
             pair<long long, pair<long long, long long>> b) {
  return a.second.second < b.second.second;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n, m;
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    par[i] = i;
    sz[i] = 1;
  }
  for (long long i = 1; i <= n; i++) {
    cin >> arr[i];
    vis[i] = 0;
  }
  cin >> m;
  long long u, v, w;
  vector<pair<long long, pair<long long, long long>>> adj(m);
  for (long long i = 1; i <= m; i++) {
    cin >> u >> v >> w;
    vis[v] = 1;
    adj[i - 1] = {u, {v, w}};
  }
  long long counter = 0;
  for (long long i = 1; i <= n; i++) {
    if (!vis[i]) {
      counter++;
      if (counter >= 2) {
        cout << -1;
        return 0;
      }
    }
  }
  sort(adj.begin(), adj.end(), compare);
  counter = 0;
  long long i = 0, ans = 0;
  memset(vis, false, sizeof(vis));
  while (counter != (n - 1)) {
    pair<long long, pair<long long, long long>> x = adj[i];
    u = find(x.first);
    v = find(x.second.first);
    if (u != v && !vis[x.second.first]) {
      vis[x.second.first] = 1;
      ans += x.second.second;
      union_(u, v);
      counter++;
    }
    i++;
  }
  cout << ans;
  return 0;
}
