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 MOD = 1000 * 1000 * 1000 + 7;
const long long INF = 1ll * 1000 * 1000 * 1000 * 1000 * 1000 * 1000 + 7;
const long long MOD2 = 998244353;
const long long N = 5000 * 100 + 1;
const long long N2 = 1000 * 1000 + 5;
const long double PI = 3.14159265;
const long long R = 250;
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x %= p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
bool sortbysec(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return (a.second < b.second);
}
const long long MAX = 1e4 + 5;
long long id[MAX], nodes, edges;
pair<long long, pair<long long, long long> > p[MAX];
long long vis[N2];
void initialize() {
  for (long long i = 0; i < MAX; ++i) id[i] = i;
}
long long root(long long x) {
  while (id[x] != x) {
    id[x] = id[id[x]];
    x = id[x];
  }
  return x;
}
void union1(long long x, long long y) {
  long long p = root(x);
  long long q = root(y);
  id[q] = id[p];
}
long long kruskal(pair<long long, pair<long long, long long> > p[]) {
  long long x, y;
  long long cost, minimumCost = 0;
  for (long long i = 0; i < edges; ++i) {
    x = p[i].second.first;
    y = p[i].second.second;
    cost = p[i].first;
    if (root(x) != root(y) && !vis[y]) {
      vis[y] = 1;
      minimumCost += cost;
      union1(x, y);
    }
  }
  return minimumCost;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long x, y;
  memset(vis, 0, sizeof(vis));
  long long weight, cost, minimumCost;
  initialize();
  cin >> nodes;
  long long q[N2];
  for (long long i = 1; i <= nodes; i++) {
    cin >> q[i];
  }
  cin >> edges;
  for (long long i = 0; i < edges; ++i) {
    cin >> x >> y >> weight;
    if (q[x] <= q[y]) continue;
    p[i] = make_pair(weight, make_pair(x, y));
  }
  sort(p, p + edges);
  minimumCost = kruskal(p);
  long long c = 0;
  for (long long i = 1; i <= nodes; i++) {
    if (id[i] == i) {
      c++;
    }
  }
  if (c == 1)
    cout << minimumCost << "\n";
  else
    cout << -1 << "\n";
}
