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.*;
import java.util.*;

public class Solution {
    
    private BufferedReader in;
    private PrintWriter out;
    private StringTokenizer st;
    
    void solve() throws IOException {
        int n = nextInt();
        for (int i = 0; i < n; ++i) {
            nextInt();
        }
        int[] min = new int[n];
        Arrays.fill(min, 1000000000);
        int m = nextInt();
        for (int i = 0; i < m; ++i) {
            int a = nextInt() - 1;
            int b = nextInt() - 1;
            int c = nextInt();
            min[b] = Math.min(min[b], c);
        }
        int ans = 0;
        boolean root = false;
        for (int i = 0; i < n; ++i) {
            if (min[i] == 1000000000) {
                if (root) {
                    out.println(-1);
                    return;
                } else {
                    root = true;
                }
            } else {
                ans += min[i];
            }
        }
        out.println(ans);
    }

    Solution() throws IOException {
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        
        eat("");
        
        solve();
        
        in.close();
        out.close();
    }
    
    private void eat(String str) {
        st = new StringTokenizer(str);
    }
    
    String next() throws IOException {
        while (!st.hasMoreTokens()) {
            String line = in.readLine();
            if (line == null) {
                return null;
            }
            eat(line);
        }
        return st.nextToken();
    }
    
    int nextInt() throws IOException {
        return Integer.parseInt(next());
    }
    
    long nextLong() throws IOException {
        return Long.parseLong(next());
    }
    
    double nextDouble() throws IOException {
        return Double.parseDouble(next());
    }

    public static void main(String[] args) throws IOException {
        new Solution();
    }
}