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>
const int N = 1e3 + 9;
using namespace std;
int n, m, cnt, fa[N];
bool vis[N];
struct Edge {
  int u, v, w;
  bool operator<(const Edge a) const { return w < a.w; }
} e[N << 4];
void add(int u, int v, int w) {
  e[cnt].u = u;
  e[cnt].v = v;
  e[cnt++].w = w;
}
int Find(int u) { return fa[u] == u ? u : fa[u] = Find(fa[u]); }
int main() {
  int u, v, w;
  scanf("%d", &n);
  for (int i = 0; i <= n; i++) scanf("%d", &m);
  for (int i = 1; i <= n; i++) fa[i] = i;
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d", &u, &v, &w);
    add(v, u, w);
  }
  sort(e, e + m);
  int ans = 0;
  for (int i = 0; i < m; i++) {
    u = e[i].u;
    v = e[i].v;
    if (vis[u]) continue;
    int fu = Find(u);
    int fv = Find(v);
    if (fu == fv) continue;
    vis[u] = 1;
    fa[fu] = fv;
    ans += e[i].w;
  }
  int sa = Find(1);
  for (int i = 2; i <= n; i++) {
    if (Find(i) != sa) return printf("-1\n"), 0;
  }
  return printf("%d\n", ans), 0;
}
