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.*;

//@author Maurice Saldivar


public class Main {

    public static void main(String[] args) throws Exception{

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pr = new PrintWriter(System.out);
        int N = Integer.parseInt(br.readLine());
        String eq[] = br.readLine().split("\\s+"); //employee qualifications
        int cost[] = new int[N + 1]; //have to make it index start at 1 other wise you will go out of range
        int M = Integer.parseInt(br.readLine());

        //populate with extremely large value ot be used as a comparison
        //if c is less than large value add it to my array
        for (int i = 0; i <= N; ++i) {
            cost[i] = Integer.MAX_VALUE;
        }

        for (int i = 0; i < M; ++i) {

            String tokens[] = br.readLine().split("\\s+");
            int a = Integer.parseInt(tokens[0]);
            int b = Integer.parseInt(tokens[1]);
            int c = Integer.parseInt(tokens[2]);

            if (c < cost[b]) {
                cost[b] = c;
            }
        }


        int sum = 0;
        int count = 0; //variable to determine if its possible to build the tree
        for (int i = 1; i <= N; ++i) {

            if (cost[i] == Integer.MAX_VALUE) {
                count++;
            } else {
                sum += cost[i];
            }
        }

        if (count == 1) {
            pr.println(sum);
            pr.flush();
        } else {            //impossible o do so print -1
            pr.println(-1);
            pr.flush();
        }
    }
}
