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;
#pragma warning(disable : 4996)
int main(int argc, char **argv) {
  const int nmax = 1000;
  const int mmax = 10000;
  int n, m, st, i, j, k, n0, err, ssc;
  unsigned int res, rest, sc[nmax], s[nmax], q[nmax], a[mmax], b[mmax], c[mmax];
  unsigned int nb[nmax], minc[nmax], nt;
  for (i = 0; i < nmax; i++) {
    q[i] = 0;
    s[i] = i;
    sc[i] = 0;
    nb[i] = 0;
    minc[i] = 0;
  }
  for (i = 0; i < mmax; i++) {
    a[i] = 0;
    b[i] = 0;
    c[i] = 0;
  }
  n = 0;
  m = 0;
  if (scanf("%d", &n) == 1) {
    for (i = 0; i < n; i++) {
      if (scanf("%u", &(q[i])) == 1) {
      }
    }
  } else {
    err = 1;
  }
  if (scanf("%d", &m) == 1) {
    for (i = 0; i < m; i++) {
      if (scanf("%u %u %u", &(a[i]), &(b[i]), &(c[i])) == 3) {
        if (nb[b[i] - 1] == 0) {
          nb[b[i] - 1]++;
          minc[b[i] - 1] = c[i];
        } else {
          if (minc[b[i] - 1] > c[i]) minc[b[i] - 1] = c[i];
        }
      }
    }
  } else {
    err = 1;
  }
  nt = 0;
  res = 0;
  for (i = 0; i < n; i++) {
    nt += nb[i];
    res += minc[i];
  }
  if (nt < n - 1) {
    printf("-1");
  } else {
    printf("%u", res);
  }
  return 0;
}
