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 Rust (1.42.0) (judge verdict: Accepted)

fn main() {
    proconio::input! {
        n: usize,
        ds: [usize; n],
        m: usize,
        ts: [usize; m],
    }
    let mut dsm = std::collections::HashMap::new();
    for d in ds {
        *dsm.entry(d).or_insert(0) += 1;
    }
    for t in ts {
        if dsm.contains_key(&t) {
            *dsm.get_mut(&t).unwrap() -= 1;
            if *dsm.get(&t).unwrap() == 0 {
                dsm.remove(&t);
            }
        } else {
            println!("NO");
            return;
        }
    }
    println!("YES");
}