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<algorithm>
int main(){
	int n;
	scanf("%d",&n);
	int d[n]={};
	for(int i=0;i<n;i++){
		scanf("%d",&d[i]);
	}
	std::sort(d,d+n);
	int m;
	scanf("%d",&m);
	int t[m];
	for(int i=0;i<m;i++){
		scanf("%d",&t[i]);
	}
	std::sort(t,t+m);
	int k=0;
	int x=0;
	int y=0;
	for(int i=0;0==0;i++){
		if(d[x]==t[y]){
			x++;
			y++;
		}else if(d[x]<t[y]){
			x++;
		}else{
			k=1;
			break;
		}
		if(y>=m)break;
	}
	if(k==1)printf("NO");
	else printf("YES");
	return 0;
}