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.util.*;
import java.io.*;

public class Main implements Runnable {

    public void solve() throws IOException {
        int n = nextInt();
        int[][] edges = new int[n][n];
        for (int i = 0; i < n; i++) {
            nextInt();
            Arrays.fill(edges[i], Integer.MAX_VALUE);
        }
        int m = nextInt();
        for (int i = 0; i < m; i++) {
            int a = nextInt() - 1;
            int b = nextInt() - 1;
            int c = nextInt();
            edges[a][b] = Math.min(edges[a][b], c);
        }
        //print(edges);
        //boolean[] done = new boolean[n];
        int tot = 0;
        int root = 0;
        for (int i = 0; i < n; i++) {
            int cost = Integer.MAX_VALUE;
            for (int j = 0; j < n; j++) {
                cost = Math.min(cost, edges[j][i]);
            }
            if (cost == Integer.MAX_VALUE) {
                root++;
                continue;
            }
            tot += cost;
        }
        if (root == 1) {
            System.out.println(tot);
        } else {
            System.out.println(-1);
        }
    }

    void shuffle(int[] a) {
        int N = a.length;
        for (int i = 0; i < N / 2; i++) {
            int j = (int) (Math.random() * N);
            int t = a[i];
            a[i] = a[j];
            a[j] = t;
        }
    }

    void print(int[] a) {
        System.out.println();
        for (int i = 0; i < a.length; i++) {
            System.out.print(a[i]);
        }
    }

    void print(int[][] arr) {
        System.out.println();
        for (int i = 0; i < arr.length; i++) {
            for (int j = 0; j < arr[i].length; j++) {
                System.out.print(arr[i][j] + " ");
            }
            System.out.println();
        }
    }

    boolean nextPermutation(int[] arr) {
        int f = -1;
        for (int i = arr.length - 2; i >= 0; i--) {
            if (arr[i + 1] > arr[i]) {
                f = i;
                break;
            }
        }
        if (f == -1) {
            return false;
        }
        int s = -1;
        for (int i = arr.length - 1; i >= 0; i--) {
            if (arr[i] > arr[f]) {
                s = i;
                break;
            }
        }
        int d = arr[f];
        arr[f] = arr[s];
        arr[s] = d;

        f++;
        int l = arr.length - 1;
        while (f < l) {
            d = arr[f];
            arr[f] = arr[l];
            arr[l] = d;
            f++;
            l--;
        }
        return true;
    }

    //-----------------------------------------------------------
    public static void main(String[] args) {
        new Main().run();
    }

    public void run() {
        try {
            in = new BufferedReader(new InputStreamReader(System.in));
            tok = null;
            solve();
            in.close();
        } catch (IOException e) {
            System.exit(0);
        }
    }

    public String nextToken() throws IOException {
        while (tok == null || !tok.hasMoreTokens()) {
            tok = new StringTokenizer(in.readLine());
        }
        return tok.nextToken();
    }

    public int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    public long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    public double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }
    BufferedReader in;
    StringTokenizer tok;
}
