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 INFI = 2000000000;
const int goX[4] = {-1, 0, 1, 0};
const int goY[4] = {0, 1, 0, -1};
const string PROBLEM_NAME = "test";
const string INPUT_FILE = PROBLEM_NAME + ".in";
const string OUTPUT_FILE = PROBLEM_NAME + ".out";
void read_input() {}
const int MAXN = 1005;
int n;
int Q[MAXN];
int a[MAXN][MAXN];
int m;
int deg[MAXN];
long long solve() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> Q[i];
  cin >> m;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) a[i][j] = INFI;
  for (int k = 0; k < m; k++) {
    int x, y, z;
    cin >> x >> y >> z;
    a[x][y] = min(a[x][y], z);
    deg[y]++;
  }
  int mcount = 0;
  for (int i = 1; i <= n; i++)
    if (deg[i] == 0) mcount++;
  if (mcount != 1) return -1;
  long long res = 0;
  for (int i = 1; i <= n; i++)
    if (deg[i]) {
      int _min = INFI + 1;
      for (int j = 1; j <= n; j++) _min = min(_min, a[j][i]);
      res += _min;
    }
  return res;
}
void m_main() { cout << solve() << endl; }
int main() {
  read_input();
  m_main();
  return 0;
}
