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.util.Arrays;
import java.util.PriorityQueue;
import java.util.Scanner;

public class Prob017B {

	static class Person implements Comparable<Person> {
		int index, quality;

		public Person(int i, int q) {
			index = i;
			quality = q;
		}

		@Override
		public int compareTo(Person arg0) {
			return quality - arg0.quality;
		}
	}

	static class App implements Comparable<App> {
		Person a, b;
		int c;

		public App(Person x, Person y, int z) {
			a = x;
			b = y;
			c = z;
		}

		@Override
		public int compareTo(App o) {
			return a.quality - o.a.quality;
		}
	}

	public static void main(String[] Args) {
		Scanner scan = new Scanner(System.in);
		int x = scan.nextInt();
		Person[] comp = new Person[x];
		for (int i = 0; i < x; i++)
			comp[i] = new Person(i, scan.nextInt());

		int apps = scan.nextInt();
		PriorityQueue<App> pq = new PriorityQueue<App>();
		for (int i = 0; i < apps; i++)
			pq.add(new App(comp[scan.nextInt() - 1], comp[scan.nextInt() - 1],
					scan.nextInt()));

		int[] costs = new int[x];
		Arrays.fill(costs, 1000001);

		while (!pq.isEmpty()) {
			App a = pq.poll();
			costs[a.b.index] = Math.min(costs[a.b.index], a.c);
		}

		int noBoss = 0;
		long sum = 0;
		for (int i = 0; i < x; i++) 
			if (costs[i] == 1000001)
				noBoss++;
			else
				sum += costs[i];

		System.out.println(noBoss == 1 ? sum : -1);
	}
}
