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.PriorityQueue;
import java.util.StringTokenizer;

import static java.lang.Integer.*;

public class DFS {
static int test;
static PriorityQueue<edge> a[];
static boolean vis[];
static long res;
    public static void main(String[] args) throws IOException {
        BufferedReader buf=new BufferedReader(new InputStreamReader(System.in));
        int n=parseInt(buf.readLine());
        buf.readLine();
        int m=parseInt(buf.readLine());
        a= new PriorityQueue[n+1];
        vis=new boolean[n+1];
        for (int i=0;i<m;i++){
            StringTokenizer st =new StringTokenizer(buf.readLine());
            int to=parseInt(st.nextToken());
            int from=parseInt(st.nextToken());
            int cost=parseInt(st.nextToken());
            if (a[from]==null)
            a[from]=new PriorityQueue<>();
            a[from].offer(new edge(to,cost));
        }
        
        for (int i=0;i<n;i++){
            dfs(i+1);
        }
        System.out.println(test!=1?-1:res);
    }
    private static void dfs(int i) {
        if (vis[i])return;
        vis[i]=true;
        if (a[i]==null || a[i].isEmpty()){
            test++;
            return;
        }
        res+=a[i].peek().r;
        dfs(a[i].peek().n);
        
    }

}
class edge implements Comparable<edge>{
    int n;
    int r;
    edge(int n,int r){
        this.n=n;
        this.r=r;
    }
    public int compareTo(edge arg0) {
        return r-arg0.r;
    }
}
