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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;

public class B {
    public static void main(String[] args) throws NumberFormatException,
            IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(in.readLine());
        String[] S = in.readLine().split(" ");
        int[] Q = new int[n];
        for (int i = 0; i < n; i++)
            Q[i] = Integer.parseInt(S[i]);

        int m = Integer.parseInt(in.readLine());
        int[][] A = new int[n][n];
        int inf = 10000000;
        for (int i = 0; i < n; i++)
            Arrays.fill(A[i], inf);
        for (int i = 0; i < m; i++) {
            S = in.readLine().split(" ");
            int x = Integer.parseInt(S[0]) - 1;
            int y = Integer.parseInt(S[1]) - 1;
            int c = Integer.parseInt(S[2]);
            A[x][y] = Math.min(A[x][y], c);
        }
        int sum = 0;
        int notDone = 0;
        boolean[] visited = new boolean[n];
        for (int j = 0; j < n; j++) {
            if (visited[j])
                continue;
            visited[j] = true;
            int min = 0;
            for (int i = 0; i < n; i++)
                if (A[i][j] < A[min][j])
                    min = i;
            if (A[min][j] == inf)
                notDone++;
            else
                sum += A[min][j];
        }
        if (notDone == 1)
            System.out.println(sum);
        else
            System.out.println(-1);
    }
}
