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.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

/**
 * Created by first on 7/26/2018.
 */
public class Hierarchy_B {
    public static void main (String [] args) throws IOException{
        Hierarchy_B x = new Hierarchy_B();
        x.solve();
    }
    public void solve() throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int numOfEmp = Integer.parseInt(br.readLine());
        int [] qualificationArr = new int[numOfEmp];
        int [] parents = new int[numOfEmp];
        int [] costArr = new int[numOfEmp];
        String in = br.readLine();
        String[] inputs = in.split(" ");
        for (int i = 0; i < numOfEmp; i++){
            qualificationArr[i] = Integer.parseInt(inputs[i]);
            parents[i] = -1;
            costArr[i] = Integer.MAX_VALUE;
        }
        int numApplications = Integer.parseInt(br.readLine());
        for(int i = 0; i < numApplications; i++){
            inputs = br.readLine().split(" ");
            int parent = Integer.parseInt(inputs[0]) - 1;
            int child = Integer.parseInt(inputs[1]) - 1;
            int cost = Integer.parseInt(inputs[2]);
            if(qualificationArr[parent] > qualificationArr[child]){
                if(cost < costArr[child]){
                    parents[child] = parent;
                    costArr[child] = cost;
                }
            }
        }
        int counter = 0;
        long result = 0;
        for (int i = 0; i < numOfEmp && counter < 2; i++){
            if(parents[i] == -1){
                counter++;
            }else {
                result += costArr[i];
            }
        }
        if(counter < 2){
            System.out.println(result);
        }else {
            System.out.println(-1);
        }
                
    }
}
