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;
int64_t modex(int64_t a, int64_t b, int64_t p) {
  int64_t res = 1;
  a %= p;
  while (b > 0) {
    if (b & 1) res = (res * a) % p;
    a = (a * a) % p;
    b >>= 1;
  }
  return res;
}
const int MD = (int)1e9 + 7;
int dx[] = {1, 0, -1, 0, 1, -1, -1, 1};
int dy[] = {0, 1, 0, -1, 1, 1, -1, -1};
int par[1003];
vector<pair<int, pair<int, int> > > g;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    par[i] = i;
    int z;
    cin >> z;
  }
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    g.push_back({c, {a, b}});
  }
  sort((g).begin(), (g).end());
  int64_t res = 0;
  for (int i = 0; i < m; i++) {
    int u = g[i].second.first;
    int v = g[i].second.second;
    if (par[v] != v) continue;
    res += g[i].first;
    par[v] = u;
  }
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    cnt += (i == par[i]);
  }
  if (cnt > 1) res = -1;
  cout << res;
  return 0;
}
