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 long long INF = 1e18;
const unsigned long long base = 2333;
const int maxn = 1e3 + 50;
const int maxm = 1e4 + 50;
const int maxv = 1e6 + 5;
const int mod = 1e9 + 7;
const int ba = 3e5;
long long a[maxn];
struct Node {
  long long u, v, w;
  bool operator<(const Node &b) const { return b.w > w; }
} n[maxm];
vector<long long> ve[maxn];
long long fa[maxn];
long long du[maxn];
long long Find(long long x) {
  if (fa[x] == x) return x;
  return fa[x] = Find(fa[x]);
}
long long sum = 0;
void dfs(long long x) {
  sum += ve[x].size();
  for (int i = 0; i < ve[x].size(); i++) dfs(ve[x][i]);
}
int main() {
  long long N;
  scanf("%lld", &N);
  for (long long i = 1; i <= N; i++) {
    scanf("%lld", &a[i]);
    fa[i] = i;
  }
  long long M;
  scanf("%lld", &M);
  for (int i = 0; i < M; i++) {
    scanf("%lld", &n[i].u);
    scanf("%lld", &n[i].v);
    scanf("%lld", &n[i].w);
    if (a[n[i].u] <= a[n[i].v]) i--, M--;
  }
  sort(n, n + M);
  long long ans = 0;
  for (int i = 0; i < M; i++) {
    long long u = n[i].u;
    long long v = n[i].v;
    long long w = n[i].w;
    long long uu = Find(u);
    long long vv = Find(v);
    if (uu == vv) continue;
    if (du[v] != 0) continue;
    du[v]++;
    fa[uu] = vv;
    ve[u].push_back(v);
    ans += w;
  }
  for (long long i = 1; i <= N; i++) {
    if (du[i] == 0) {
      dfs(i);
      break;
    }
  }
  if (sum == N - 1)
    printf("%lld\n", ans);
  else
    printf("-1\n");
  return 0;
}
