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.*;

class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = Integer.parseInt(sc.next());
        HashMap<Integer, Integer> h = new HashMap<>();
        for (int i = 0; i < n; i++) {
            int d = Integer.parseInt(sc.next());
            if (h.containsKey(d)) {
                h.put(d, h.get(d) + 1);
            } else {
                h.put(d, 1);
            }
        }
        int m = Integer.parseInt(sc.next());
        HashMap<Integer, Integer> g = new HashMap<>();
        for (int i = 0; i < m; i++) {
            int t = Integer.parseInt(sc.next());
            if (g.containsKey(t)) {
                g.put(t, g.get(t) + 1);
            } else {
                g.put(t, 1);
            }
        }
        sc.close();
        boolean f = true;
        for (int k : g.keySet()) {
            if (!h.containsKey(k) || h.get(k) < g.get(k)) {
                f = false;
                break;
            }
        }
        System.out.println((f ? "YES" : "NO"));
    }
}