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 <algorithm>
#include <iostream>
#include <math.h>
#include <vector>
using namespace std;

int main(){
	int n;
	cin >> n;
	vector<int> d;
	for (int i = 0; i < n; i++) {
		int tmp;
		cin >> tmp;
		d.push_back(tmp);
	}
	
	sort(d.begin(), d.end());

	int m;
	bool flag;
	cin >> m;
	vector<int>::iterator t;
	for(int i = 0; i < m; i++) {
		int tmp;
		cin >> tmp;
		if (binary_search(d.begin(), d.end(), tmp)) {
			t = find(d.begin(), d.end(), tmp);
			d.erase(t);
			n--;

		} else {
			cout << "NO";
			return 0;
		}

	}

	cout << "YES";	

	return 0;
}