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.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Scanner;

/*
 * To change this template, choose Tools | Templates
 * and open the template in the editor.
 */
/**
 *
 * @author pttrung
 */
public class B {

    public static void main(String[] args) {
        Scanner in = new Scanner(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);
        int n = in.nextInt();
        int[] cost = new int[n];
        int[] q = new int[n];
        Arrays.fill(cost, -1);
        for (int i = 0; i < n; i++) {
            q[i] = in.nextInt();
        }
        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();
            if (q[a] > q[b]) {
                if (cost[b] == -1 || cost[b] > c) {
                    cost[b] = c;
                }
            }
        }
        int total = 0;
        int count = 0;
        int max = 0;
        for (int i = 0; i < n; i++) {
            if (cost[i] == -1) {
                count++;
            } else {
                total += cost[i];
                max = Math.max(cost[i], max);
            }

        }
        if (count > 1) {
            out.println(-1);
        } else {
            if (count == 0) {
                out.println(total - max);
            }else{
                out.println(total);
            }
        }
        out.close();

    }

    public double pow(double a, int b) {
        if (b == 0) {
            return 1;
        }
        if (b == 1) {
            return a;
        }
        double val = pow(a, b / 2);
        if (b % 2 == 0) {
            return val * val;
        } else {
            return val * val * a;
        }
    }

    public static long gcd(long a, long b) {
        if (b == 0) {
            return a;
        }
        return gcd(b, a % b);
    }

    public static long lcm(long a, long b) {
        return a * b / gcd(a, b);
    }
}
