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.InputStreamReader;
import java.io.StreamTokenizer;
import java.util.Arrays;


public class B {

	public static void main(String[] args) throws Exception {
		new B().run();
	}
	
	class Item implements Comparable<Item> {
		int p;
		int index;
		Item(int p, int index) {
			this.p = p;
			this.index = index;
		}
		@Override
		public int compareTo(Item o) {
			return -(p - o.p);
		}
	}

	private StreamTokenizer st;
	
	int e[], next[], f[], w[], rc;

	private void run() throws Exception {
		st = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
		int n = nextInt();
		//Item[] items = new Item[n];
		int[] q = new int[n];
		int fi = -1;
		for (int i = 0; i < n; i++) {
			q[i] = nextInt();
			if (fi == -1 || q[i] > q[fi]) {
				fi = i;
			}
		}
		int m = nextInt();
		f = new int[n];
		Arrays.fill(f, -1);
		e = new int[m];
		next = new int[m];
		w = new int[m];
		rc = 0;
		for (int i = 0; i < m; i++) {
			int u = nextInt() - 1;
			int v = nextInt() - 1;
			int c = nextInt();
			add(u, v, c);
		}
		int[] cost = new int[n];
		Arrays.fill(cost, Integer.MAX_VALUE / 2);
		cost[fi] = 0;
		int[] col = new int[n];
		Arrays.fill(col, 0);
		while (true) {
			int min = Integer.MAX_VALUE / 2;
			int ti = -1;
			for (int i = 0; i < n; i++) {
				if (cost[i] < min && col[i] == 0) {
					min = cost[i];
					ti = i;
				}
			}
			if (ti == -1) {
				break;
			}
			col[ti] = 1;
			int j = f[ti];
			while (j > -1) {
				cost[e[j]] = Math.min(cost[e[j]], w[j]);
				j = next[j];
			}
		}
		
		boolean flag = true;
		int res = 0;
		for (int i = 0; i < n; i++) {
			if (cost[i] == Integer.MAX_VALUE / 2)
				flag = false;
			res += cost[i];
		}
		
		System.out.println(flag ? res : -1);
	}
	
	private void add(int u, int v, int c) {
		e[rc] = v;
		next[rc] = f[u];
		f[u] = rc;
		w[rc] = c;
		rc++;
	}

	private int nextInt() throws Exception {
		st.nextToken();
		return (int) st.nval;
	}

}
