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.PriorityQueue;
import java.util.Scanner;


public class Main {
    public static void main(String[] args) {
        Scanner r = new Scanner(System.in);
        
        int n = r.nextInt();

        int[] q = new int[n];
        for(int i = 0; i < n; i++)
            q[i] = r.nextInt();
        
        
        PriorityQueue<Edge> pq = new PriorityQueue<Edge>();
        int m = r.nextInt();
        for(int k = 0; k < m; k++){
            int i = r.nextInt()-1;
            int j = r.nextInt()-1;
            int w = r.nextInt();
            
            pq.add(new Edge(i, j, w));
        }
        
        boolean[] t = new boolean[n];
        int[] p = new int[n];
        int cc = n;
        for(int i = 0; i < n; i++)
            p[i] = i;
        
        int cost = 0;
        while(!pq.isEmpty()){
            Edge e = pq.poll();
            
            if(!isSame(e.i, e.j, p) && !t[e.j]){
                union(e.i, e.j, p);
                t[e.j] = true;
                cost += e.w;
                cc--;
            }
        }
        
        
        if(cc == 1)System.out.println(cost);
        else System.out.println(-1);
    }
    static int find(int i, int[] p){
        if(i == p[i])return i;
        else return p[i] = find(p[i], p);
    }
    static void union(int i, int j, int[] p){
        p[find(i, p)] = find(j, p);
    }
    static boolean isSame(int i, int j, int[] p){
        return find(i, p) == find(j, p);
    }
}
class Edge implements Comparable<Edge>{
    int i, j, w;
    public Edge(int ii, int ji, int wi){
        i = ii;
        j = ji;
        w = wi;
    }
    @Override
    public int compareTo(Edge b) {
        if(w < b.w)return -1;
        else return 1;
    }
}