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.
/*
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 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */


import java.util.Scanner;

/**
 *
 * @author magdy
 */
public class Super {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
       
    
      
        Scanner input = new Scanner(System.in);
        int x = input.nextInt();
        int a[] = new int [x];
        for (int i = 0; i < x; i++) {
            a[i] = input.nextInt();
        }
        int y = input.nextInt();
        int [][] b = new int[y][3];
        for(int i=0 ; i<y ; i++){
            b[i][0] = input.nextInt()-1;
            b[i][1] = input.nextInt()-1;
            b[i][2] = input.nextInt();
        }
        
        int [] dsu = new int[x];
        for(int i=0 ; i<x ; i++) dsu[i] =i;
        
        int [] cost = new int[x];
        for(int i=0 ; i<y ;i++){
           if( dsu[b[i][1]] == b[i][1] || cost[b[i][1]] > b[i][2]){
               union(dsu,b[i][0],b[i][1]);
               cost[b[i][1]] = b[i][2];
           }
        }
        int ans =0;
        for(int i=0 ; i<x ;i++){
            ans+= cost[i];
        }
        if(num(dsu)){
        System.out.println(ans);
        }else{
            System.out.println("-1");
        }
    }
    
    
    
     static void union(int [] arr , int a,int b){
        int root_a = root(arr,a);
        int root_b = root(arr,b);
        arr[b] = root_a;
       
    }
    
     static int root (int [] arr , int i){
        while(arr[i] != i) {
            arr[i] = arr[arr[i]];
            i = arr[i];
        }
        return i;
        
    }
    
     static boolean find(int []arr,int A,int B)
{
    if( root(arr,A)==root(arr,B) )       //if A and B have same root,means they are connected.
    return true;
    else
    return false;
}
    
    static boolean num(int [] arr){
       int x = arr[0];
       for (int i = 0; i < arr.length-1; i++) {
           if( root(arr,i) != root(arr,i+1)) return false;
       }
       return true;
   }
    
   
}
