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 MAX = 1e3 + 3;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
int n, m;
int q[MAX];
bool taken[MAX];
int p[MAX], rnk[MAX];
vector<pair<int, pair<int, int> > > edg;
void init() {
  for (int i = 0; i < MAX; i++) {
    p[i] = i;
    rnk[i] = 0;
  }
  memset(taken, 0, sizeof taken);
  edg.clear();
}
int findSet(int i) { return (i == p[i] ? i : p[i] = findSet(p[i])); }
void unionSet(int i, int j) {
  int x = findSet(i), y = findSet(j);
  if (x != y) {
    if (rnk[x] > rnk[y])
      p[y] = x;
    else {
      p[x] = y;
      if (rnk[x] == rnk[y]) rnk[y]++;
    }
  }
}
int main() {
  init();
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", q + i);
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    int x, y, w;
    scanf("%d %d %d", &x, &y, &w);
    edg.push_back({w, {x, y}});
  }
  sort(edg.begin(), edg.end());
  int cost = 0;
  for (int i = 0; i < m; i++) {
    int x = edg[i].second.first;
    int y = edg[i].second.second;
    int w = edg[i].first;
    if (findSet(x) != findSet(y) && q[x] > q[y] && !taken[y]) {
      unionSet(x, y);
      cost += w;
      taken[y] = 1;
    }
  }
  int cnt = 0;
  for (int i = 1; i <= n; i++)
    if (findSet(i) == i) cnt++;
  if (cnt > 1)
    puts("-1");
  else
    printf("%d\n", cost);
}
