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 INF = 1e9;
const long long INF64 = 4e18;
const int M = 2 * 100 + 10;
const int N = 2 * 100 * 1000 + 100;
const int MOD = 1e9 + 7;
const double EPS = 1e-9;
const double PI = 3.14159265359;
long long gcd(long long a, long long b) { return a == 0 ? b : gcd(b % a, a); }
int p[N];
int s[N];
int used[N];
int get(int x) {
  if (p[x] == x) return x;
  return p[x] = get(p[x]);
}
void merge(int x, int y) {
  x = get(x);
  y = get(y);
  if (x == y) return;
  if (s[x] < s[y]) swap(x, y);
  p[y] = x;
  s[x] += s[y];
}
void init(int n) {
  for (int i = 0; i < int(n); ++i) {
    p[i] = i;
    s[i] = 1;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> q(n);
  for (int i = 0; i < int(n); ++i) {
    cin >> q[i];
  }
  int m;
  cin >> m;
  vector<pair<int, pair<int, int>>> e(m);
  for (int i = 0; i < int(m); ++i) {
    int c, v, u;
    cin >> v >> u >> c;
    v--;
    u--;
    e[i] = {c, {v, u}};
  }
  sort(e.begin(), e.end());
  init(n);
  int cost = 0;
  int k = 0;
  for (int i = 0; i < int(m); ++i) {
    int c = e[i].first, v = e[i].second.first, u = e[i].second.second;
    if (get(v) != get(u) && !used[u]) {
      used[u] = true;
      merge(v, u);
      cost += c;
      k++;
    }
  }
  cout << (k != n - 1 ? -1 : cost) << '\n';
}
