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 = 10005;
struct edge {
  int u, v, c, pre;
} E[N * 2];
int head[N], cnt;
void add_edge(int u, int v, int c) {
  E[cnt].u = u;
  E[cnt].v = v;
  E[cnt].c = c;
  E[cnt].pre = head[u];
  head[u] = cnt++;
}
int res;
bool vis[N / 10 + 5];
int Set[1005];
int Find_Set(int x) {
  if (Set[x] == x) return x;
  return Set[x] = Find_Set(Set[x]);
}
void Union_Set(int x, int y) { Set[Find_Set(x)] = Find_Set(y); }
void dfs(int u) {
  int small = 10000000, pos = -1;
  for (int i = head[u]; i != -1; i = E[i].pre) {
    int v = E[i].v;
    int c = E[i].c;
    if (c < small) {
      small = c;
      pos = i;
    }
  }
  if (pos != -1) {
    Union_Set(E[pos].v, u);
    res += small;
  }
}
int val[N / 10 + 5];
int main() {
  int n;
  while (~scanf("%d", &n)) {
    memset(head, -1, sizeof(head));
    memset(vis, false, sizeof(vis));
    for (int i = 1; i <= n; i++) Set[i] = i;
    cnt = 0;
    for (int i = 1; i <= n; i++) scanf("%d", val + i);
    int m;
    scanf("%d", &m);
    while (m--) {
      int u, v, c;
      scanf("%d%d%d", &u, &v, &c);
      if (v != u) add_edge(v, u, c);
    }
    res = 0;
    for (int i = 1; i <= n; i++) {
      dfs(i);
    }
    set<int> q;
    q.clear();
    for (int i = 1; i <= n; i++) q.insert(Set[Find_Set(i)]);
    if (q.size() > 1)
      printf("%d\n", -1);
    else
      printf("%d\n", res);
  }
  return 0;
}
