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.*;

public class b {
	public static int[][] graph;
	public static int[] qual;
	public static int n;
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		n = in.nextInt();
		graph = new int[n][n];
		qual = new int[n];
		emp[] e = new emp[n];
		for(int i=0; i<n; i++) {
			qual[i] = in.nextInt();
			e[i] = new emp(qual[i],i);
		}
		Arrays.sort(e);
		int m = in.nextInt();
		for(int i=0; i<n; i++)
			Arrays.fill(graph[i],-1);
		for(int i=0; i<m; i++) {
			int a = in.nextInt()-1;
			int b = in.nextInt()-1;
			int c = in.nextInt();
			if(qual[a] <= qual[b]) 
				continue;
			if(graph[a][b] == -1)
				graph[a][b] = c;
			graph[a][b] = Math.min(graph[a][b],c);
		}
		long sum = 0;
		boolean[] found = new boolean[n];
		found[e[0].p] = true;
		for(int i=1; i<n; i++) {
			int lookfor = e[i].p;

			long min = (long)1e10;
			for(int j=0; j<n; j++) {
				if(graph[j][lookfor] != -1 && found[j]) {
					if(graph[j][lookfor] < min)
						min = graph[j][lookfor];
				}
			}
			sum += min;

			found[e[i].p] = true;
		}
		if(sum >= (long)1e10)
			System.out.println(-1);
		else
			System.out.println(sum);
	}
	public static boolean bfs(int start) {
		boolean[] visited = new boolean[n];
		Queue<Integer> q = new LinkedList<Integer>();
		q.add(start);
		visited[start] = true;
		while(!q.isEmpty()) {
			int crnt = q.poll();
			for(int i=0; i<n; i++) {
				if(graph[crnt][i] != -1 && !visited[i]) {
					visited[i] = true;
					q.add(i);
				}
			}
		}
		for(int i=0; i<n; i++)
			if(!visited[i])
				return false;
		return true;
	}
}
class emp implements Comparable<emp> {
	int q, p;
	emp(int a, int b) {
		q = a;
		p = b;
	}
	public int compareTo(emp e) {
		if(e.q-q != 0)
			return e.q-q;
		return p-e.p;
	}
}
class Edge implements Comparable<Edge> {
	int to, w;
	Edge(int a, int b) {
		to = a;
		w = b;
	}
	public int compareTo(Edge e) {
		return w-e.w;
	}
}
