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;
import java.util.ArrayList;
import java.util.TreeMap;

/**
 * Created by Chris on 3/23/2015.
 */
public class Hierarchy {


    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));


        int minCost = 0;

        int n = Integer.parseInt(br.readLine());
        String[] line = br.readLine().split(" ");

        int[] qual = new int[n + 1];
        ArrayList<TreeMap<Integer, Integer>> hierarchy = new ArrayList<TreeMap<Integer, Integer>>();
        int maxQual = -1;
        int boxIndex = -1;
        int tmp;
        for (int i = 1; i <= n; i++) {
            tmp = Integer.parseInt(line[i - 1]);
            qual[i] = tmp;
            hierarchy.add(i-1, new TreeMap<Integer, Integer>());
            if (tmp > maxQual) {
                maxQual = tmp;
                boxIndex = i;
            } else if (tmp == maxQual) {
                System.out.println(-1);
                return;
            }

        }

        int m = Integer.parseInt(br.readLine());


        // hierarchy[i] - A map from cost to index, ordered by cost, the index can supervise i

        int a, b, c;
        for (int i = 0; i < m; i++) {
            line = br.readLine().split(" ");
            a = Integer.parseInt(line[0]); //supervisor
            b = Integer.parseInt(line[1]); //subordinate
            c = Integer.parseInt(line[2]); //cost
            hierarchy.get(b - 1).put(c, a - 1);
        }

        int countWithouthSupervisor = 0;
        for(TreeMap<Integer, Integer> map : hierarchy)
        {

            if(map.isEmpty())
            {
                countWithouthSupervisor++;
                if(countWithouthSupervisor > 1)
                {
                    System.out.println(-1);
                    return;
                }
            }
            else
            {
                minCost += map.firstKey();
            }
        }
        System.out.println(minCost);
    }
}
