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.
//package com.company;
import java.io.*;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader buffer = new BufferedReader(new InputStreamReader(System.in));
        int empl = Integer.parseInt(buffer.readLine());

        //shift the array to allign the employee indexes
        int[] mvs = new int[empl + 1];
        for (int i = 1; i <= empl; i++)
            mvs[i] = Integer.MAX_VALUE;
        buffer.readLine(); //dont need to know the quals

        //a is ready to be a supervisor over b at cost c
        //record the cost in the max values integer array
        int applications = Integer.parseInt(buffer.readLine());
        for (int i = 0; i < applications; i++) {
            String[] line = buffer.readLine().split(" ");
            int employeea = Integer.parseInt(line[0]);
            int employeeb = Integer.parseInt(line[1]);
            int cost = Integer.parseInt(line[2]);
            if (mvs[employeeb] > cost)
                mvs[employeeb] = cost;
        }

        //count the cost and track who didn't get get a sup.
        int ans = 0;
        int count = 0;
        for (int i = 1; i < empl + 1; i++)
            if (mvs[i] == Integer.MAX_VALUE){
                count++;
            }else {
                ans = ans + mvs[i];
            }
        if(count == 1){ //make sure one does NOT have a supervisor otherwise can't create tree
            System.out.println(ans);
        }else {
            System.out.println("-1");
        }
    }
}