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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#define datatype int

//比較関数
int compare(datatype a, datatype b){
	return a - b;
}

//昇順
void sort_sub(datatype *origin, int left, int right, datatype *tmp){
	if(right - left > 1){
		int i, j, k, half = (left + right) / 2;
		sort_sub(origin, left, half, tmp);
		sort_sub(origin, half, right, tmp);
		for(i = left; i < right; i++){
			tmp[i] = origin[i];
		}
		for(i = left, j = left, k = half; i < right; i++){
			if(k == right){
				origin[i] = tmp[j];
				j++;
			}
			else if(compare(tmp[j], tmp[k]) <= 0 && j < half){
				origin[i] = tmp[j];
				j++;
			}
			else{
				origin[i] = tmp[k];
				k++;
			}
		}
	}
}

void sort(datatype *origin, int N){
	datatype *tmp = (datatype *)malloc(sizeof(datatype) * N);
	sort_sub(origin, 0, N, tmp);
	free(tmp);
}

int main(){
	int N, M, i, j;
	scanf("%d", &N);
	int *D = (int *)malloc(sizeof(int) * N);
	for(i = 0; i < N; i++){
		scanf("%d", &D[i]);
	}
	scanf("%d", &M);
	int *T = (int *)malloc(sizeof(int) * M);
	for(j = 0; j < M; j++){
		scanf("%d", &T[j]);
	}
	sort(D, N);
	sort(T, M);
	for(i = 0, j = 0; j < M && i < N;){
		if(D[i] == T[j]){
			i++;
			j++;
		}
		else if(D[i] > T[j]){
			printf("NO\n");
			return 0;
		}
		else{
			i++;
		}
	}
	if(j < M){
		printf("NO\n");
	}
	else{
		printf("YES\n");
	}
	return 0;
}