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.File;
import java.io.FileReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Locale;
import java.util.StringTokenizer;

public class Solution implements Runnable {
	
	public static void main(String[] args) {
		(new Thread(new Solution())).start();
	}
	
	BufferedReader in;
	PrintWriter out;
	StringTokenizer st;
	
	String nextToken() throws IOException {
		while (st == null || !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());
	}
	
	int[] q;
	
	class R implements Comparable<R>{
		int a, b, c;

		public int compareTo(R o) {
			if (q[b] == q[o.b]) return c - o.c; else return q[o.b] - q[b];
		}
	}
	
	int[] dsu;
	int[] r;
	
	int get(int a) {
		if (dsu[a] == a) return a; else return dsu[a] = get(dsu[a]);
	}
	
	void uni(int a, int b) {
		a = get(a); b = get(b);
		if (r[a] == r[b]) r[a]++;
		if (r[a] > r[b]) dsu[b] = a; else dsu[a] = b;
	}
	
	void solve() throws Exception {
		int n = nextInt();
		dsu = new int[n + 1];
		r = new int[n + 1];
		for (int i = 0; i <= n; i++) {
			dsu[i] = i;
			r[i] = 0;
		}
		q = new int[n + 1];
		for (int i = 1; i <= n; i++) q[i] = nextInt();
		int m = nextInt();
		R[] rr = new R[m];
		for (int i = 0; i < m; i++) {
			rr[i] = new R();
			rr[i].a = nextInt();
			rr[i].b = nextInt();
			rr[i].c = nextInt();
		}
		Arrays.sort(rr);
		int[] d = new int[n + 1];
		long ans = 0;
		for (int i = 0; i < m; i++) {
			if (d[rr[i].b] == 0) {
				d[rr[i].b] = rr[i].a;
				ans += rr[i].c;
			}
		}
		int j = 0;
		for (int i = 1; i <= n; i++) {
			if (d[i] == 0) 
				if (j != 0) ans = -1; else j = i;
		}
		out.println(ans);
	}
	
	public void run() {
		try {
			Locale.setDefault(Locale.US);
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
			solve();
		} catch (Exception e) {
			e.printStackTrace();
		}
		out.flush();
	}

}
