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;
const long long M = 1000000007;
long long parent[1010];
long long s[1010];
bool supervisor[1010];
vector<pair<long long, pair<long long, long long>>> g;
long long find(long long e) {
  long long root = e;
  while (root != parent[root]) {
    root = parent[root];
  }
  while (e != root) {
    long long p = parent[e];
    parent[e] = root;
    e = p;
  }
  return root;
}
bool doUnion(long long a, long long b) {
  long long roota = find(a);
  long long rootb = find(b);
  if (roota == rootb || supervisor[b]) return false;
  if (s[roota] < s[rootb]) {
    parent[roota] = rootb;
    s[rootb]++;
  } else {
    parent[rootb] = roota;
    s[roota]++;
  }
  return true;
}
long long kruskal() {
  int x, y;
  long long cost, minimumCost = 0;
  for (int i = 0; i < g.size(); ++i) {
    x = g[i].second.first;
    y = g[i].second.second;
    cost = g[i].first;
    if (doUnion(x, y)) {
      minimumCost += cost;
      supervisor[y] = true;
    }
  }
  return minimumCost;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  g.clear();
  long long a[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    parent[i] = i;
    s[i] = 1;
    supervisor[i] = false;
  }
  long long m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    long long s, t, c;
    cin >> s >> t >> c;
    s--, t--;
    g.push_back({c, {s, t}});
  }
  sort(g.begin(), g.end());
  long long ans = kruskal();
  long long c = 0;
  for (int i = 0; i < n; i++)
    if (!supervisor[i]) {
      c++;
    }
  if (c > 1) ans = -1;
  if (n == 1) ans = 0;
  cout << ans << "\n";
  return 0;
}
