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.*;

public class Solution {

	Scanner in;
	PrintWriter out;

	class Edge implements Comparable<Edge> {
		int u, v, w;
		@Override
		public int compareTo(Edge e) {
			return w - e.w;
		}
	}
	
	int N;
	Edge E[];

	int findSet(int u) {
		if (P[u] == u) return u;
		return P[u] = findSet(P[u]);
	}

	void union_(int u, int v) {
		link(findSet(u), findSet(v));
	}

	void link(int u, int v) {
		if (R[u] < R[v]) {
			P[u] = v;
		} else {
			P[v] = u;
			if (R[u] == R[v]) ++R[u];
		}
	}

	int P[], R[];

	void solve() throws IOException {
		in = new Scanner(System.in);
		//in = new Scanner(new FileReader("input.txt"));
		out = new PrintWriter(System.out);
		N = in.nextInt();
		P = new int[N];
		R = new int[N];
		for (int i = 0; i < N; ++i) {
			in.nextInt();
			P[i] = i;
		}
		int M = in.nextInt();
		E = new Edge[M];
		for (int i = 0; i < M; ++i) {
			E[i] = new Edge();
			E[i].u = in.nextInt() - 1;
			E[i].v = in.nextInt() - 1;
			E[i].w = in.nextInt();
		}
		int a[] = new int[N];
		int b[] = new int[N];
		Arrays.fill(a, -1);
		Arrays.fill(b, -1);
		Arrays.sort(E);

		int ans = 0;
		int c = 0;
		for (int i = 0; i < M; ++i) {
			int u = E[i].u;
			int v = E[i].v;
			int w = E[i].w;
			if (a[v] == -1) {
				a[v] = u;
				b[v] = w;
			} else if (b[v] < c) {
				b[v] = w;
				a[v] = u;
			}
			//if (findSet(u) == findSet(v)) continue;
			//union_(u, v);
			//ans += w;
			//++c;
		}
		for (int i = 0; i < N; ++i) {
			if (a[i] != -1) {
				ans += b[i];
				c++;
			}
		}
		if (c != N - 1) ans = -1;
		out.println(ans);
		out.close();		
	}

	public static void main(String args[]) throws IOException {
		new Solution().solve();
	}
}
