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: Wrong Answer)

import java.util.*;

public class Main {
	public static void main (String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int[] arrD = new int[n];
		for (int i = 0; i < n; i++) {
			arrD[i] = sc.nextInt();
		}
		int m = sc.nextInt();
		int[] arrT = new int[m];
		for (int i = 0; i < m; i++) {
			arrT[i] = sc.nextInt();
		}
		Arrays.sort(arrD);
		Arrays.sort(arrT);
		int idx = 0;
		for (int i = 0; i < m; i++) {
			while (true) {
				if (idx >= n || arrD[idx] > arrT[i]) {
					System.out.println("No");
					return;
				} else if (arrD[idx] == arrT[i]) {
					idx++;
					break;
				} else {
					idx++;
				}
			}
		}
		System.out.println("Yes");
	}
}