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 A[1001][1001], D[1001];
vector<int> W[1001];
int q[1001], cnt, N, M, i, j, mini, v, a, b, c, t;
bool l[1001][1001];
bool f[1001];
queue<int> w;
bool bfs(int k) {
  for (i = 1; i <= N; i++) f[i] = false;
  w.push(k);
  f[k] = true;
  cnt = 0;
  while (!w.empty()) {
    k = w.front();
    w.pop();
    cnt++;
    for (j = 0; j < W[k].size(); j++)
      if (!f[W[k][j]]) {
        w.push(W[k][j]);
        f[W[k][j]] = true;
      }
  }
  if (cnt == N) return true;
  return false;
}
long long Prim(int t) {
  for (i = 1; i <= N; i++) {
    f[i] = false;
    D[i] = 100000000;
  }
  D[t] = 0;
  for (i = 1; i <= N; i++) {
    mini = 100000000;
    for (j = 1; j <= N; j++)
      if (!f[j] && D[j] < mini) {
        mini = D[j];
        v = j;
      }
    f[v] = true;
    for (j = 0; j < W[v].size(); j++)
      if (D[W[v][j]] > A[v][W[v][j]]) D[W[v][j]] = A[v][W[v][j]];
  }
  long long sum = 0;
  for (i = 1; i <= N; i++) sum += D[i];
  return sum;
}
int main(int argc, char *argv[]) {
  cin >> N;
  for (i = 1; i <= N; i++) cin >> q[i];
  cin >> M;
  for (i = 1; i <= M; i++) {
    cin >> a >> b >> c;
    if (l[a][b]) {
      if (c < A[a][b]) A[a][b] = c;
    } else if (q[a] > q[b]) {
      W[a].push_back(b);
      A[a][b] = c;
      f[b] = true;
      l[a][b] = true;
    }
  }
  for (i = 1; i <= N; i++)
    if (!f[i]) break;
  if (i > N) {
    cout << -1 << endl;
    return 0;
  }
  t = i;
  if (bfs(i))
    cout << Prim(t) << endl;
  else
    cout << -1 << endl;
  return 0;
}
