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 B {
    static class Edge implements Comparable<Edge>{
        int a,b,cost;
        public Edge(int aa,int bb,int e){
            a=aa;
            b=bb;
            cost=e;
        }
        public int compareTo(Edge e){
            if(e.cost<cost) return 1;
            return -1;
        }
    }
    static int parent[];
    public static void union(int x,int y){
        int px=find(x);
        int py=find(y);
        if(px==py) return;
        else parent[px]=py;
    }
    public static int find(int x){
        if(parent[x]==x) return parent[x];
        return parent[x]=find(parent[x]);
    }
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        int[] qual=new int[n];
        for(int i=0;i<n;i++)
            qual[i]=sc.nextInt();
        int app=sc.nextInt();
        PriorityQueue<Edge>pq=new PriorityQueue<Edge>();
        int[][] cost=new int[n][n];
        for(int i=0;i<n;i++) Arrays.fill(cost[i], 10000000);
        for(int i=0;i<app;i++){
            int st=sc.nextInt()-1;
            int end=sc.nextInt()-1;
            cost[st][end]=Math.min(cost[st][end], sc.nextInt());
        }
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                if(cost[i][j]!=10000000)
                    pq.add(new Edge(i,j,cost[i][j]));
            }
        }
        parent=new int[n];
        for(int i=0;i<n;i++) parent[i]=i;
        int count=0;
        int res=0;
        int[] countSuperVisors=new int[n];
        while(!pq.isEmpty()&&count<n-1){
            Edge tmp=pq.remove();
            if(find(tmp.a)==find(tmp.b)||countSuperVisors[tmp.b]>0) continue;
            count++;
            res+=tmp.cost;
            union(tmp.a,tmp.b);
            countSuperVisors[tmp.b]++;
        }
        if(count<n-1) System.out.println(-1);
        else System.out.println(res);
        
    }
}
