Problem:
Score : 200 points

Problem Statement

Rng is preparing a problem set for a qualification round of CODEFESTIVAL.

He has N candidates of problems. The difficulty of the i-th candidate is D_i.

There must be M problems in the problem set, and the difficulty of the i-th problem must be T_i. Here, one candidate of a problem cannot be used as multiple problems.

Determine whether Rng can complete the problem set without creating new candidates of problems.

Constraints

1 \leq N \leq 200,000

1 \leq D_i \leq 10^9

1 \leq M \leq 200,000

1 \leq T_i \leq 10^9

All numbers in the input are integers.

Partial Score

100 points will be awarded for passing the test set satisfying N \leq 100 and M \leq 100.

Input

Input is given from Standard Input in the following format:

N
D_1 D_2 ... D_N
M
T_1 T_2 ... T_M

Output

Print YES if Rng can complete the problem set without creating new candidates of problems; print NO if he cannot.

Sample Input 1

5
3 1 4 1 5
3
5 4 3

Sample Output 1

YES

Sample Input 2

7
100 200 500 700 1200 1600 2000
6
100 200 500 700 1600 1600

Sample Output 2

NO

Not enough 1600s.

Sample Input 3

1
800
5
100 100 100 100 100

Sample Output 3

NO

Sample Input 4

15
1 2 2 3 3 3 4 4 4 4 5 5 5 5 5
9
5 4 3 2 1 2 3 4 5

Sample Output 4

YES

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Arrays;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner scanner = null;
		try {
			scanner = new Scanner(System.in);
			execute(scanner);
		} finally {
			if (scanner != null) {
				scanner.close();
			}
		}
	}

	private static void execute(Scanner scanner) {
		int n = scanner.nextInt();
		int[] d = new int[n];
		for (int i = 0; i < n; i++) {
			d[i] = scanner.nextInt();
		}
		int m = scanner.nextInt();
		int[] t = new int[m];
		for (int i = 0; i < m; i++) {
			t[i] = scanner.nextInt();
		}
		Arrays.sort(d);
		Arrays.sort(t);
		int j = 0;
		for (int i = 0; i < m; i++) {
			while (j < n && t[i] != d[j]) {
				j++;
			}
			if (j < n && t[i] == d[j]) {
				j++;
				continue;
			}
			if (j == n) {
				System.out.println("NO");
				return;
			}
		}
		System.out.println("YES");
	}
}