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;
int pset[1010];
void initset() {
  for (int i = 0, _n = 1010; i < _n; i++) pset[i] = i;
}
int findSet(int i) { return (pset[i] == i) ? i : (pset[i] = findSet(pset[i])); }
void merge(int i, int j) { pset[findSet(i)] = findSet(j); }
int sameSet(int i, int j) { return findSet(i) == findSet(j); }
int main() {
  for (int n, m, a, b, c; scanf("%d", &n) != EOF;) {
    vector<int> q(n), p(n, -1);
    for (int i = 0, _n = n; i < _n; i++) scanf("%d", &q[i]);
    scanf("%d", &m);
    vector<pair<int, pair<int, int> > > E;
    for (int i = 0, _n = m; i < _n; i++) {
      scanf("%d %d %d", &a, &b, &c);
      a--;
      b--;
      E.push_back(make_pair(c, make_pair(a, b)));
    }
    sort(E.begin(), E.end());
    int cost = 0;
    initset();
    for (int i = 0, _n = m; i < _n; i++) {
      c = E[i].first;
      a = E[i].second.first;
      b = E[i].second.second;
      if (!sameSet(a, b) && p[b] == -1) {
        p[b] = a;
        merge(a, b);
        cost += c;
      }
    }
    bool ok = true;
    for (int i = 0, _n = n - 1; i < _n; i++)
      if (!sameSet(0, i + 1)) ok = false;
    if (!ok)
      puts("-1");
    else
      printf("%d\n", cost);
  }
}
