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 = 1001;
int a[N];
int vis[N];
int n;
struct edge {
  int from, to, w;
  bool operator<(const edge& e) const { return w > e.w; }
};
vector<edge> adj[N];
priority_queue<edge> q;
int MST() {
  int mstCost = 0;
  int edges = 0;
  while (!q.empty()) {
    edge e = q.top();
    q.pop();
    if (vis[e.to]) continue;
    vis[e.to] = 1;
    mstCost += e.w;
    ++edges;
    for (int j = 0; j < adj[e.to].size(); j++) {
      edge ne = adj[e.to][j];
      if (!vis[ne.to]) q.push(ne);
    }
  }
  if (edges != n - 1) return -1;
  return mstCost;
}
int main() {
  int m;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    int u, v, w;
    scanf("%d %d %d", &u, &v, &w);
    --u, --v;
    adj[u].push_back(edge{u, v, w});
    q.push(edge{u, v, w});
  }
  printf("%d\n", MST());
}
