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 int N = 2e5 + 69;
const int64_t INF = 100000000000000018LL;
long long p[N], n, szz[N], m, k;
void make(long long n) {
  for (long long i = 1; i <= n; i++) {
    p[i] = i;
    szz[i] = 1;
  }
  return;
}
long long find(long long x) {
  if (p[x] == x) return x;
  return p[x] = find(p[x]);
}
long long merge(long long a, long long b) {
  long long x = find(a);
  long long y = find(b);
  if (x != y) {
    if (szz[x] < szz[y]) swap(x, y);
    p[y] = x;
    szz[x] += szz[y];
    return 1;
  }
  return 0;
}
void solve() {
  long long n;
  cin >> n;
  long long a[n + 1], i;
  for (i = 1; i <= n; i++) cin >> a[i];
  vector<pair<long long, pair<long long, long long>>> v;
  long long m, x, y, z;
  cin >> m;
  for (i = 1; i <= m; i++) {
    cin >> x >> y >> z;
    v.push_back({z, {x, y}});
  }
  make(n);
  sort((v).begin(), (v).end());
  set<long long> s;
  long long ans = 0, cnt = 0;
  for (i = 0; i < (long long)v.size(); i++) {
    x = v[i].second.first;
    y = v[i].second.second;
    z = v[i].first;
    if (cnt != n - 1) {
      if (a[x] > a[y]) {
        if (find(x) != find(y) && s.count(y) == 0) {
          merge(x, y);
          s.insert(y);
          ans += z;
          cnt++;
        }
      }
    }
  }
  if (cnt != n - 1)
    cout << -1;
  else
    cout << ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
}
