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.*;
import java.math.*;
import java.util.regex.*;
import java.lang.*;

/**
 *
 * @author jon
 */
public class Round17B {
    int cost[][];// = new int[1001][1001];
    Ppl qual[];// = new int[1001];
    int oo = (int)1e10;
    Round17B() throws Exception{
        Scanner in = new Scanner(System.in);
        int N = in.nextInt();
        qual = new Ppl[N];
        cost = new int[N][N];
        for(int i = 0; i < N; i++) Arrays.fill(cost[i], oo);
        for(int i = 0; i < N; i++) {
            qual[i] = new Ppl(i, in.nextInt());
        }

        Arrays.sort(qual);

        int M = in.nextInt();
        for(int i = 0; i < M; i++) {
            int a = in.nextInt()-1;
            int b = in.nextInt()-1;
            int c = in.nextInt();

            cost[a][b] = Math.min(cost[a][b], c);
        }
        int ret = 0;
        TreeSet<Integer> used = new TreeSet<Integer>();
        used.add(qual[0].ind);
        for(int i = 1; i < qual.length; i++) {
            int temp = oo;
            for(int a : used) {
                temp = Math.min(temp, cost[a][qual[i].ind]);
//                System.out.println(a + " " + qual[i].ind);
            }
            if(temp == oo) {
//                System.out.println(i);
                ret = oo;
                break;
            } else {
                ret += temp;
                used.add(qual[i].ind);
            }
        }

        if(ret == oo) System.out.println("-1");
        else System.out.println(ret);

    }
    public class Ppl implements Comparable<Ppl> {
        int qual;
        int ind;
        Ppl(int a, int b) {
            ind = a;
            qual = b;
        }
        public int compareTo(Ppl b) {
            return b.qual - this.qual;
        }
    }
    public static void main(String args[]) throws Exception {
        new Round17B();
    }
}
