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.HashMap;
import java.util.Map;
import java.util.Scanner;


	public class Main {

		public static void main(String[] args) {

			Scanner sc = new Scanner(System.in);

			int n = sc.nextInt();
			int[] d = new int[n];
			int maxd = 0;
			for(int i=0;i<n;i++) {
				d[i] = sc.nextInt();
				maxd = Math.max(maxd, d[i]);
			}
			
			int m = sc.nextInt();
			int[] t = new int[m];
			int maxt = 0;
			for(int i=0;i<m;i++) {
				t[i]= sc.nextInt();
				maxt = Math.max(maxt, t[i]);
			}
			
			int maxi = Math.max(maxd, maxt);
			
			Map<Integer,Integer> difd = new HashMap<Integer,Integer>(); 
			
			
			for(int i=0;i<n;i++) {
				if(!difd.containsKey(d[i])) {
					difd.put(d[i], 1);
				}else {
					difd.put(d[i], difd.get(d[i])+1);
				}
			}
			
			for(int i=0;i<m;i++) {
				if(!(difd.containsKey(t[i]))){
					System.out.println("NO");
					return;
				}
				if(!(difd.get(t[i])>0)){
					System.out.println("NO");
					return;
				}
				
				difd.put(t[i], difd.get(t[i])-1);
			}
			
			System.out.println("YES");
			
		}
		
		
		
	}