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 MAXN = 1e3 + 5;
int n, m;
int d[MAXN];
set<pair<int, int> > heap;
vector<pair<int, int> > adj[MAXN];
int s, nmax;
void dijk() {
  for (int i = 1; i <= n; ++i) d[i] = 1e9;
  d[s] = 0;
  heap.insert(pair<int, int>(d[s], s));
  while (!heap.empty()) {
    pair<int, int> p = *heap.begin();
    int u = p.second;
    d[u] = p.first;
    heap.erase(heap.begin());
    for (int i = 0; i <= (int)adj[u].size() - 1; ++i) {
      int v = adj[u][i].second;
      if (adj[u][i].first < d[v]) {
        std::set<pair<int, int> >::iterator it =
            heap.find(pair<int, int>(d[v], v));
        if (it != heap.end()) heap.erase(it);
        d[v] = adj[u][i].first;
        heap.insert(pair<int, int>(d[v], v));
      }
    }
  }
  int res = 0;
  for (int i = 1; i <= n; ++i) {
    if (d[i] == 1e9) {
      printf("-1");
      return;
    }
    res += d[i];
  }
  printf("%d", res);
}
int main() {
  scanf("%d", &n);
  nmax = 0;
  for (int i = 1; i <= n; ++i) {
    int qu;
    scanf("%d", &qu);
    if (qu > nmax) {
      nmax = qu;
      s = i;
    }
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; ++i) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    adj[u].push_back(pair<int, int>(w, v));
  }
  dijk();
  return 0;
}
