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: Accepted)

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#include<queue>
using namespace std;
int main(void){
	vector<int>a;
	vector<int>b;
	int d;
	int k;
	int n, m;
	int s, t;
	int x, y, z;

	scanf("%d", &n);
	for (int i = 0; i < n; i++){
		scanf("%d", &x);
		a.push_back(x);
	}
	scanf("%d", &m);
	for (int i = 0; i < m; i++){
		scanf("%d", &x);
		b.push_back(x);
	}
	sort(a.begin(), a.end());
	sort(b.begin(), b.end());
	s = 0;
	t = 0;
	for (; s < n&&t < m;){
		if (a[s] == b[t]){
			s++;
			t++;
		}
		else s++;
	}
	if(t==m)printf("YES\n");
	else printf("NO\n");

	return 0;
}