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 maxn = 1000 + 10;
const long long INF = 999999999999ll;
int n, back[maxn];
long long dis[maxn];
vector<pair<long long, int> > e[maxn];
bool mark[maxn][maxn];
bool cmp(pair<long long, int> a, pair<long long, int> b) {
  return a.second < b.second;
}
int prime(int source) {
  for (int i = 0; i < maxn; i++) dis[i] = INF;
  set<pair<long long, int> > st;
  st.clear();
  st.insert(make_pair(0, source));
  dis[source] = 0;
  long long ret = 0;
  while (!st.empty()) {
    int cur = (*st.begin()).second;
    st.erase(st.begin());
    for (int i = 0; i < e[cur].size(); i++)
      if (dis[e[cur][i].first] > e[cur][i].second &&
          !mark[cur][e[cur][i].first]) {
        mark[cur][e[cur][i].first] = true;
        ret -= back[e[cur][i].first];
        ret += e[cur][i].second;
        back[e[cur][i].first] = e[cur][i].second;
        if (st.find(make_pair(dis[e[cur][i].first], e[cur][i].first)) !=
            st.end())
          st.erase(st.find(make_pair(dis[e[cur][i].first], e[cur][i].first)));
        dis[e[cur][i].first] = e[cur][i].second;
        st.insert(make_pair(dis[e[cur][i].first], e[cur][i].first));
      }
  }
  bool correct = true;
  for (int i = 1; i <= n; i++)
    if (dis[i] == INF) {
      correct = false;
      break;
    }
  if (!correct) ret = -1;
  return ret;
}
int main() {
  long long an;
  pair<long long, int> source;
  cin >> n;
  for (int i = 1; i <= n; i++)
    cin >> an, source = max(source, make_pair(an, i));
  int m, fi, se, w;
  cin >> m;
  while (m-- > 0) cin >> fi >> se >> w, e[fi].push_back(make_pair(se, w));
  for (int i = 1; i <= n; i++) sort(e[i].begin(), e[i].end(), cmp);
  cout << prime(source.second) << endl;
  return 0;
}
