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;
vector<long long> p;
vector<long long> sz;
long long get_root(long long v) {
  if (p[v] == -1) {
    return v;
  }
  long long root = get_root(p[v]);
  p[v] = root;
  return root;
}
void unite(long long a, long long b) {
  a = get_root(a);
  b = get_root(b);
  if (a == b) {
    return;
  }
  if (sz[a] < sz[b]) {
    p[a] = b;
    sz[b] += sz[a];
  } else {
    p[b] = a;
    sz[a] += sz[b];
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  long long n, m, a, b, c;
  cin >> n;
  vector<vector<long long> > g;
  p.assign(n, -1);
  sz.assign(n, 1);
  vector<long long> n1(n, 0);
  vector<long long> f(n);
  for (long long i = 0; i < n; i++) cin >> f[i];
  cin >> m;
  for (long long i = 0; i < m; i++) {
    cin >> a >> b >> c;
    a--;
    b--;
    g.push_back({c, a, b});
  }
  sort(g.begin(), g.end());
  long long comp = n;
  long long ans = 0;
  for (auto i : g) {
    if (get_root(i[1]) == get_root(i[2]) || n1[i[2]] != 0 || comp == 1) {
      continue;
    }
    unite(i[1], i[2]);
    comp--;
    n1[i[2]] = 1;
    ans += i[0];
  }
  if (comp == 1)
    cout << ans;
  else
    cout << -1;
  return 0;
}
