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.*;
import java.util.*;


public class hierarchy {
    
    static int [] parent;
    static int [] maxValue;
    static int [] rank;
    public static int find(int num)
    {
        if (num != parent[num])
            parent[num] = find(parent[num]);
        return parent[num];
    }
    
    public static void main (String args[]) throws IOException
    {
        //BufferedReader cin  = new BufferedReader(new FileReader("c:\\input.txt"));
        BufferedReader cin = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter cout = new PrintWriter(System.out);
        int n = Integer.parseInt(cin.readLine());
        int [] e = new int [n];
        String [] s = cin.readLine().split(" ");
        for (int i = 0; i < n; i++)
        {
            e[i] = Integer.parseInt(s[i]);
        }
        int m = Integer.parseInt(cin.readLine());
        if (m < n - 1)
            cout.println("-1");
        else
        {
            int totalCost = 0;
            parent = new int [n + 1];
            //rank = new int [n];
            for (int i = 0; i < n + 1; i++)
            {
                parent[i] = i;
            }
            Application [] apps = new Application [m];
            for (int i = 0; i < m; i++)
            {
                Application app = new Application();
                s = cin.readLine().split(" ");
                app.a = Integer.parseInt(s[0]);
                app.b = Integer.parseInt(s[1]);
                app.c = Integer.parseInt(s[2]);
                apps[i] = app;
            }
            Arrays.sort(apps);
            for (int i = 0; i < m; i++)
            {
                int a = apps[i].a;
                int b = apps[i].b;
                int c = apps[i].c;
                int setA = find(a);
                int setB = find(b);
                if (setA == setB) //already in the same set
                {
                    
                }
                else if (setB == b)//b is the head of its set so we can join
                {
                    parent[setB] = parent[setA];
                    totalCost += c;
                }
            }
            int u = find(1);
            boolean possible = true;
            for (int i = 2; i < n + 1; i++)
            {
                int u2 = find(i);
                if (u != u2)
                {
                    cout.println("-1");
                    possible = false;
                    break;
                }
            }
            if (possible)
                cout.println(totalCost);
        }
        cout.close();
    }
}

class Application implements Comparable<Application>
{
    public int a;
    public int b;
    public int c;
    @Override
    public int compareTo(Application app) {
        if (c < app.c)
            return - 1;
        else if (c > app.c)
            return 1;
        else
            return 0;
    }
}
