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;
int n, m;
int par[1001];
priority_queue<pair<int, pair<int, int> >, vector<pair<int, pair<int, int> > >,
               greater<pair<int, pair<int, int> > > >
    pq;
bool hasSuper[1001], visited[1001];
vector<int> g[1001];
int findRepresentative(int x) {
  if (par[x] == x) {
    return x;
  } else {
    par[x] = findRepresentative(par[x]);
    return par[x];
  }
}
void dfs(int x) {
  if (!visited[x]) {
    visited[x] = true;
    for (int i : g[x]) {
      dfs(i);
    }
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    par[i] = i;
    hasSuper[i] = false;
  }
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    pq.push(make_pair(c, make_pair(a, b)));
  }
  long long int ans = 0;
  while (!pq.empty()) {
    int nowCost = pq.top().first;
    pair<int, int> node = pq.top().second;
    int nodeA = node.first;
    int nodeB = node.second;
    pq.pop();
    int parA = findRepresentative(nodeA);
    int parB = findRepresentative(nodeB);
    if (parA != parB && !hasSuper[nodeB]) {
      par[parA] = parB;
      ans += nowCost;
      hasSuper[nodeB] = true;
      g[nodeA].push_back(nodeB);
      g[nodeB].push_back(nodeA);
    }
  }
  dfs(1);
  for (int i = 1; i <= n; i++) {
    if (!visited[i]) {
      cout << -1 << endl;
      return 0;
    }
  }
  cout << ans << endl;
}
