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.
//package codeforces.br17;

import java.io.*;

/**
 * User: Kandy
 * Date: 10.06.2010
 * Time: 20:36:09
 */
public class ProblemB {
    public void solve() {
        boolean oj = true;

        try {
            Reader reader = oj ? new InputStreamReader(System.in) : new FileReader("B.in");
            Writer writer = oj ? new OutputStreamWriter(System.out) : new FileWriter("B.out");
            BufferedReader br = new BufferedReader(reader);
            PrintWriter out = new PrintWriter(writer);

            MyTokenizer tok = new MyTokenizer(br.readLine());
            int n = (int)tok.getNum();
            tok = new MyTokenizer(br.readLine());
            int[] q = new int[n];
            for(int i=0;i<n;i++)
                q[i] = (int)tok.getNum();
            tok = new MyTokenizer(br.readLine());
            int m = (int)tok.getNum();
            int[][] r = new int[m][3];

            int[] min = new int[n];
            for(int i=0;i<n;i++) {
                min[i] = Integer.MAX_VALUE;
            }

            for(int i=0;i<m;i++) {
                tok = new MyTokenizer(br.readLine());
                r[i][0] = (int)tok.getNum() - 1;
                r[i][1] = (int)tok.getNum() - 1;
                r[i][2] = (int)tok.getNum();
                min[r[i][1]] = Math.min(min[r[i][1]], r[i][2]);
            }

            int count = 0;
            long sum = 0;
            for(int i=0;i<n;i++)
                if (min[i] == Integer.MAX_VALUE)
                    count++;
                else
                    sum += min[i];

            if (count == 1) {
                out.printf("%d", sum);
            } else {
                out.printf("-1");
            }

            br.close();
            out.close();
            reader.close();
            writer.close();
        }
        catch (Exception ex) {
            ex.printStackTrace();
        }
        finally {
        }
    }

    public static void main(String[] args) {
        ProblemB f = new ProblemB();
        f.solve();
    }

    private class MyTokenizer {
        private String s;
        private int cur;

        public MyTokenizer(String s) {
            this.s = s;
            cur = 0;
        }

        public void skip() {
            while (cur < s.length() && (s.charAt(cur) == ' ' || s.charAt(cur) == '\n')) {
                cur++;
            }
        }

        public double getNum() {
            skip();
            String snum = "";
            while (cur < s.length() && (s.charAt(cur) >= '0' && s.charAt(cur) <= '9' || s.charAt(cur) == '.')) {
                snum += s.charAt(cur);
                cur++;
            }
            return Double.valueOf(snum);
        }

        public String getString() {
            skip();
            String s2 = "";
            while (cur < s.length() && (s.charAt(cur) >= 'a' && s.charAt(cur) <= 'z')) {
                s2 += s.charAt(cur);
                cur++;
            }
            return s2;
        }

        public char getCurrentChar() throws Exception {
            if (cur < s.length())
                return s.charAt(cur);
            else
                throw new Exception("Current character out of string length");
        }

        public void moveNextChar() {
            if (cur < s.length())
                cur++;
        }

        public boolean isFinished() {
            return cur >= s.length();
        }
    }
}
