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.*;
import java.util.*;
import static java.lang.Math.*;
import static java.util.Arrays.fill;
import static java.util.Arrays.binarySearch;
import static java.util.Arrays.sort;

public class Main {
	public static void main(String[] args) throws IOException {
		try {
			if (new File("input.txt").exists()) {
				System.setIn(new FileInputStream("input.txt"));
			}
		} catch (SecurityException e) {
		}
		
		new Main().run();
	}

	BufferedReader in;
	PrintWriter out;
	StringTokenizer st = new StringTokenizer("");
	
	int vNum;
	int eNum;
	int[] q;
	Edge[] edges;
	boolean[] used;
	
	void run() throws IOException {
		in = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		
		vNum = nextInt();
		q = new int [vNum];
		
		for (int i = 0; i < vNum; i++)
			q[i] = nextInt();
		
		eNum = nextInt();
		edges = new Edge [eNum];
		
		for (int i = 0; i < eNum; i++)
			edges[i] = new Edge(nextInt() - 1, nextInt() - 1, nextInt());
		
		sort(edges);
		DSF dsf = new DSF(vNum);
		
		long ans = 0L;
		used = new boolean [vNum];
		
		for (Edge e : edges) {
			if (!used[e.v] && dsf.union(e.u, e.v)) {
				used[e.v] = true;
				ans += e.w;
			}
		}
		
		out.println(dsf.sscCount() == 1 ? ans : -1);
		
		out.close();
	}
	
	class Edge implements Comparable<Edge> {
		int u;
		int v;
		long w;
		
		Edge(int u, int v, long w) {
			this.u = u;
			this.v = v;
			this.w = w;
		}
		
		@Override
		public int compareTo(Edge e) {
			return w == e.w ? 0 : (w < e.w ? -1 : 1);
		}
	}
	
	class DSF {
		int[] set;
		int[] rnk;
		
		DSF(int sz) {
			set = new int [sz];
			rnk = new int [sz];
			for (int i = 0; i < sz; i++)
				set[i] = i;
		}
		
		int setOf(int i) {
			return i == set[i] ? i : (set[i] = setOf(set[i]));
		}
		
		boolean union(int i, int j) {
			i = setOf(i);
			j = setOf(j);
			if (i == j)
				return false;
			if (rnk[i] > rnk[j]) {
				set[j] = i;
			} else {
				set[i] = j;
				if (rnk[i] == rnk[j])
					rnk[j]++;
			}
			return true;
		}
		
		int sscCount() {
			int res = 0;
			for (int i = 0; i < set.length; i++) {
				int s = setOf(i);
				if (rnk[s] >= 0) {
					rnk[s] = -1;
					res++;
				}
			}
			return res;
		}
	}
	
	String nextToken() throws IOException {
		while (!st.hasMoreTokens()) {
			st = new StringTokenizer(in.readLine());
		}
		
		return st.nextToken();
	}
	
	int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}
	
	long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}
	
	double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}
	
	String nextLine() throws IOException {
		st = new StringTokenizer("");
		return in.readLine();
	}
	
	boolean EOF() throws IOException {
		while (!st.hasMoreTokens()) {
			String s = in.readLine();
			if (s == null)
				return true;
			st = new StringTokenizer(s);
		}
		return false;
	}
}
