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 MAX = LLONG_MAX, MOD = 1000000007, MODA = 1e9 - 1;
const long long MIN = LLONG_MIN;
const long double PI = 3.1415926535;
const long long N = 100009;
bool visited[N] = {0};
vector<long long> adj[N];
void dfs(long long node) {
  if (!visited[node]) {
    visited[node] = 1;
    for (auto u : adj[node]) {
      dfs(u);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  while (t--) {
    long long n;
    cin >> n;
    vector<pair<long long, long long>> v;
    for (long long i = 0; i < n; i++) {
      long long x;
      cin >> x;
      v.push_back({x, i + 1});
    }
    sort(v.begin(), v.end());
    vector<pair<long long, long long>> master[n + 1];
    long long m;
    cin >> m;
    for (long long i = 0; i < m; i++) {
      long long a, b, c;
      cin >> a >> b >> c;
      master[b].push_back({c, a});
    }
    long long ans = 0;
    for (long long i = 0; i < n - 1; i++) {
      if (master[v[i].second].size() == 0) break;
      sort(master[v[i].second].begin(), master[v[i].second].end());
      ans += master[v[i].second][0].first;
      adj[v[i].second].push_back(master[v[i].second][0].second);
      adj[master[v[i].second][0].second].push_back(v[i].second);
    }
    dfs(v[n - 1].second);
    bool flag = 0;
    for (long long i = 1; i < n + 1; i++) {
      if (!visited[i]) {
        flag = 1;
        break;
      }
    }
    if (flag)
      cout << -1;
    else
      cout << ans;
  }
  return 0;
}
