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 C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
#define lli long long int
#define OK -1
#define NO -2
#define USED -3



int main () {

	lli n,m;
	lli *d,*t;

	cin >> n;
	d = new lli[n];
	for(lli i=0;i<n;i++){
		cin >> d[i];
	}

	sort(d, d+n);

	cin >> m;
	t = new lli[m];
	for(lli i=0;i<m;i++){
		cin >> t[i];
	}
	sort(t, t+m);


	for(lli i=0;i<m;i++){
		for(lli j=0;j<n;j++){
			if(t[i]==d[j]){
				t[i] = OK;
				d[j] = USED;
			}else if(t[i]<d[j]){
				break;
			}
		}
		if(t[i]!=OK){
			cout << "NO" << endl;
			return 0;
		}
	}

	cout << "YES" << endl;

	

    return 0;
}