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.ArrayList;
import java.util.List;
import java.util.Scanner;

class Main {

    public static void main(String[] args) {
        new Main().compute();
    }

    void compute() {
        Scanner sc = new Scanner(System.in);
        long N = sc.nextLong();
        List<Long> D = new ArrayList<>();
        for (int i = 0; i < N; i++) {
            D.add(sc.nextLong());
        }
        D.sort((o1, o2) -> (int) (o1 - o2));
        Long M = sc.nextLong();
        List<Long> T = new ArrayList<>();
        for (int i = 0; i < M; i++) {
            T.add(sc.nextLong());
        }
        T.sort((o1, o2) -> (int) (o1 - o2));
        int current=0;
        check:for(int i=0;i<M;i++){
            for(int j=current;j<N;j++){
                if(T.get(i)==D.get(j)){
                    current+=1;
                    continue check;
                }
                if(j==N-1){
                    System.out.println("No");
                    return;
                }
            }
        }
        System.out.println("Yes");
    }
}