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.util.*;
import java.lang.*;
import java.io.*;

public class Main
{
    public static void main (String[] args) throws java.lang.Exception
    {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int N = Integer.parseInt(br.readLine());
        String[] quals = br.readLine().split(" ");
        
        int[] cost = new int[N+1];
        for(int i=0; i<=N;i++) {
            cost[i] = Integer.MAX_VALUE;
        }
        //each node (except one) has to be added as a subordinate exactly once.
        //just find the minimum offer to supervise each person and make sure exactly
        //one isn't supervised. It will always be the one with the highest quals, so there
        //can't be an offer to supervise everyone.
        int M = Integer.parseInt(br.readLine());
        for(int i=0; i<M;i++) {
            String[] app = br.readLine().split(" ");
            int a, b, c;
            a = Integer.parseInt(app[0]);
            b = Integer.parseInt(app[1]);
            c = Integer.parseInt(app[2]);
            if(c < cost[b]) {
                cost[b] = c;
            }
        }
        
        int sum = 0;
        int count = 0;
        for(int i=1; i<=N; i++) {
            //System.out.println("cost @ "+i+" = "+cost[i]);
            if(cost[i] == Integer.MAX_VALUE) {
                count++;
            }
            else {
                sum += cost[i];
            }
            
        }
        
        if(count == 1) {
            System.out.println(sum);
        }
        else { System.out.println(-1); }
    }
}