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.lang.*;
import java.io.*;
import java.util.*;
import java.math.*;

public class Solution implements Runnable{
    private static BufferedReader br = null;
    private static PrintWriter out = null;
    private static StringTokenizer stk = null;
    
    public static void main(String[] args) {
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        (new Thread(new Solution())).start();
    }
    
    private void loadLine() {
        try {
            stk = new StringTokenizer(br.readLine());
        }
        catch (IOException e) {
            e.printStackTrace();
        }
    }
    
    private String nextLine() {
        try {
            return br.readLine();
        }
        catch (IOException e) {
            e.printStackTrace();
        }
        return null;
    }
    
    private Integer nextInt() {
        while (stk==null||!stk.hasMoreTokens()) loadLine();
        return Integer.parseInt(stk.nextToken());
    }
    
    private Long nextLong() {
        while (stk==null||!stk.hasMoreTokens()) loadLine();
        return Long.parseLong(stk.nextToken());
    }
    
    private String nextWord() {
        while (stk==null||!stk.hasMoreTokens()) loadLine();
        return (stk.nextToken());
    }
    
    private Double nextDouble() {
        while (stk==null||!stk.hasMoreTokens()) loadLine();
        return Double.parseDouble(stk.nextToken());
    }
    
    class E implements Comparable<E> {
        int a,b;
        public E(int a, int b) {
            this.a=a;
            this.b=b;
        }
        
        public int compareTo(E a) {
            return this.b==a.b?this.a-a.a:-this.b+a.b;
        }
    }
    
    public void run() {
        int n = nextInt();
        E[] arr = new E[n];
        for (int i = 0; i < n; ++i) {
            arr[i] = new E(i, nextInt());
        }
        
        Arrays.sort(arr);
        int m = nextInt();
        
        int[][] z = new int[m][3];
        for (int i = 0; i < m; ++i) {
            z[i][0] = nextInt()-1;
            z[i][1] = nextInt()-1;
            z[i][2] = nextInt();
        }
        
        boolean[] con = new boolean[n];
        Arrays.fill(con, false);
        con[arr[0].a] = true;
        
        int res = 0;
        for (int i = 1; i < n; ++i) {
            int min = Integer.MAX_VALUE;
            int idx = -1;
            for (int j = 0; j < m; ++j) {
                if (z[j][1] == arr[i].a) {
                    min = Math.min(min, z[j][2]);
                    idx = j;
                }
            }
            
            if (idx == -1) {
                res = -1;
                break;
            }
            else {
                res += min;
                con[arr[i].a] |= con[z[idx][0]];
            }
        }
        
        for (int i = 0; i < n; ++i) {
            if (!con[i]) {
                res =-1;
            }
        }
        
        out.println(res);
        out.flush();
    }
}

