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;
long long graph[1002][1002];
long long q[1002];
vector<pair<long long, int> > sortedq;
long long int mincost[1002];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &q[i]);
    sortedq.push_back(make_pair(-q[i], i));
  }
  sort(sortedq.begin(), sortedq.end());
  int m;
  scanf("%d", &m);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) graph[i][j] = 10000000000000000ll;
  for (int i = 1; i <= m; i++) {
    int from, to;
    long long cost;
    scanf("%d %d %lld", &from, &to, &cost);
    graph[from][to] = min(graph[from][to], cost);
  }
  mincost[0] = 0;
  for (int i = 1; i < n; i++) {
    mincost[i] = 10000000000000000ll;
    int to = sortedq[i].second;
    int qto = -sortedq[i].first;
    for (int j = 0; j < i; j++) {
      int from = sortedq[j].second;
      int qfrom = -sortedq[j].first;
      if (qfrom > qto) {
        mincost[i] = min(mincost[i], mincost[i - 1] + graph[from][to]);
      }
    }
  }
  if (mincost[n - 1] == 10000000000000000ll)
    printf("-1\n");
  else
    printf("%lld\n", mincost[n - 1]);
}
