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;
struct Edge {
  int u;
  int v;
  int w;
  Edge() { u = v = w = -1; }
  Edge(int a, int b, int c) : u(a), v(b), w(c) {}
};
int N, M, qa, qb, qc;
vector<pair<int, int> > adj[1000 + 5];
bool vis[1000 + 5];
int D[1000 + 5];
int PAR[1000 + 5];
int ST[1000 + 5];
int stIdx;
vector<Edge> Edges;
bool hasIn[1000 + 5];
int inWeight[1000 + 5];
void dfs(int v) {
  vis[v] = true;
  const vector<pair<int, int> >& adjVec = adj[v];
  int len = (int)adjVec.size();
  int nextV;
  for (int i = 0; i < len; i++) {
    nextV = adjVec[i].first;
    if (!vis[nextV]) {
      dfs(nextV);
    }
  }
  ST[stIdx++] = v;
}
int dfsTwo(int v) {
  vis[v] = true;
  int sum = 1, nextV;
  const vector<pair<int, int> >& adjVec = adj[v];
  int len = (int)adjVec.size();
  for (int i = 0; i < len; i++) {
    nextV = adjVec[i].first;
    if (!vis[nextV]) {
      sum += dfsTwo(nextV);
    }
  }
  return sum;
}
int main() {
  int tmpInt = 0, root = 0, v, d, nextV, nextD, len, cost;
  scanf("%d", &N);
  for (int i = 0; i < N; i++) {
    scanf("%d", &tmpInt);
  }
  scanf("%d", &M);
  for (int i = 0; i < M; i++) {
    scanf("%d %d %d", &qa, &qb, &qc);
    adj[qa].push_back(pair<int, int>(qb, qc));
    Edges.push_back(Edge(qa, qb, qc));
  }
  for (int i = 1; i <= N; i++) {
    if (!vis[i]) {
      dfs(i);
    }
  }
  root = ST[stIdx - 1];
  memset(vis, 0, sizeof(vis));
  if (N != dfsTwo(root)) {
    printf("-1\n");
  } else {
    cost = 0;
    for (int i = 0; i < M; i++) {
      const Edge& edge = Edges[i];
      v = edge.v;
      if (!hasIn[v]) {
        hasIn[v] = true;
        inWeight[v] = edge.w;
        cost += edge.w;
      } else if (edge.w < inWeight[v]) {
        cost = cost - inWeight[v] + edge.w;
        inWeight[v] = edge.w;
      }
    }
    printf("%d\n", cost);
  }
  return 0;
}
