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.*;
public class Main {
	public static void main(String args[]) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int D[] = new int[N];
		for(int i = 0; i < N; i++) {
			D[i] = sc.nextInt();
		}
		int M = sc.nextInt();
		int T[] = new int[M];
		for(int i = 0; i < M; i++) {
			T[i] = sc.nextInt();
		}
		Arrays.sort(D);
		Arrays.sort(T);
		int cnt = 0;
		out : for(int i = 0; i < M; i++) {
			for(int j = cnt; j < N; j++) {
				if(T[i] == D[j]) {
					if(i == M - 1) {
						System.out.println("YES");
						return;
					} else {
						cnt = j + 1;
						continue out;
					}
				}
			}
			System.out.println("NO");
			return;
		}
	}
}