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.Arrays;
import java.util.Scanner;

public class B {
    static class Sort implements Comparable<Sort> {
        int a,b,c;
        public int compareTo(Sort argo) {
            return this.c-argo.c;
        }
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[]q = new int[n+1];
        for (int i = 1; i <= n; i++) {
            q[i] = sc.nextInt();
        }
        int maxind = 1;
        for (int i = 2; i <= n; i++) {
            if (q[i] > q[maxind])
                maxind = i;
        }
        for (int i = 1; i <= n; i++) {
            if (i != maxind && q[i]==q[maxind]) {
                System.out.println(-1);
                return;
            }
        }
        int m = sc.nextInt();
        Sort[]costs = new Sort[m];
        for (int i = 0; i < m; i++) {
            costs[i] = new Sort();
            costs[i].a = sc.nextInt();
            costs[i].b = sc.nextInt();
            costs[i].c = sc.nextInt();
        }
        Arrays.sort(costs);
        int[]p = new int[n+1];
        int ans = 0, cnt = 1;
        p[maxind] = maxind;
        for (int i = 0; i < m; i++) {
            if (p[costs[i].b]==0 || p[costs[i].b]==costs[i].b) {
                if (p[costs[i].a]==0) {
                    p[costs[i].a] = costs[i].a;
                    cnt++;
                }
                if (p[costs[i].b]==0)
                    cnt++;
                p[costs[i].b] = p[costs[i].a];
                for (int j = 1; j <= n; j++) {
                    if (p[j]==costs[i].b)
                        p[j] = p[costs[i].a];
                }
                ans += costs[i].c;
            }
        }
        boolean f = true;
        for (int i = 1; i <= n; i++) {
            if (p[i]!=maxind) {
                f = false;
                break;
            }
        }
        if (!f)
            ans = -1;
        if (cnt != n)
            ans = -1;
        System.out.println(ans);
    }
}
