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.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.List;

public class ProblemB {
	
	static class Edge implements Comparable<Edge> {
		int a;
		int b;
		long cost;
		Edge(int _a, int _b, long _c) {
			a = _a;
			b = _b;
			cost = _c;
		}
		@Override
		public int compareTo(Edge arg0) {
			return Long.signum(cost - arg0.cost);
		}
	}
	
    static class UnionFind {
        int[] parent, rank;
        UnionFind(int n) {
             parent = new int[n];
             rank = new int[n];
             for (int i = 0 ; i < n ; i++) {
                  parent[i] = i;
                  rank[i] = 0;
             }
        }
       
        int find(int x) {
             if (parent[x] == x) {
                  return x;
             }
             parent[x] = find(parent[x]);
             return parent[x];
        }
       
        void unite(int x, int y) {
             x = find(x);
             y = find(y);
             if (x == y) {
                  return;
             }
             if (rank[x] < rank[y]) {
                  parent[x] = y;
             } else {
                  parent[y] = x;
                  if (rank[x] == rank[y]) {
                       rank[x]++;
                  }
             }
        }
        boolean issame(int x, int y) {
             return (find(x) == find(y));
        }
   }

	
	public static void main(String[] args) throws IOException {
		BufferedReader s = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(System.out);

		int n = Integer.valueOf(s.readLine());
		String[] qd = s.readLine().split(" ");
		int[][] q = new int[n][2];
		for (int i = 0 ; i < n ; i++) {
			q[i][0] = Integer.valueOf(qd[i]);
			q[i][1] = i;
		}
		Arrays.sort(q, new Comparator<int[]>(){
			@Override
			public int compare(int[] arg0, int[] arg1) {
				return arg1[0] - arg0[0];
			}
		});
		
		
		int m = Integer.valueOf(s.readLine());
		List<Edge> e = new ArrayList<Edge>();
//		List<Edge>[] graph = new List[n];
//		for (int i = 0 ; i < n ; i++) {
//			graph[i] = new ArrayList<Edge>();
//		}
		for (int i = 0 ; i < m ; i++) {
			String[] data = s.readLine().split(" ");
			int a = Integer.valueOf(data[0]) - 1;
			int b = Integer.valueOf(data[1]) - 1;
			int c = Integer.valueOf(data[2]);
			e.add(new Edge(a, b, c));
		}
		Collections.shuffle(e);
		Collections.sort(e);

		long totalCost = 0;
		int addedEdge = 0;
		boolean[] in = new boolean[n];
		UnionFind uf = new UnionFind(n);
		for (Edge edge : e) {
			if (uf.issame(edge.a, edge.b) || in[edge.b]) {
				continue;
			}
			in[edge.b] = true;
			uf.unite(edge.a, edge.b);
			addedEdge++;
			totalCost += edge.cost;
 		}
		
		if (addedEdge == n-1) {
			out.println(totalCost);
		} else {
			out.println(-1);
		}
		out.flush();
	}

	public static void debug(Object... os){
		System.err.println(Arrays.deepToString(os));
	}
}