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>
#include <math.h>
#include <string.h>
#include <limits.h>

int cmp(const void* p, const void* q) {
	return *(int*)q - *(int*)p;
}

int main(void) {
	int n, m;
	int d[200000], t[200000];
	int flag = 0, buf = 0;

	scanf("%d", &n);
	for (int i = 0; i < n; i++) scanf("%d", &d[i]);
	scanf("%d", &m);
	for (int i = 0; i < m; i++) scanf("%d", &t[i]);

	qsort(d, n, sizeof(int), cmp);
	qsort(t, m, sizeof(int), cmp);
	for (int i = 0; i < m; i++) {
		for (int j = buf; j < n; j++) {
			if (t[i] == d[j]) {
				buf = j + 1, flag++;
				break;
			}
		}
	}
	//printf("flag=%d\n", flag);
	if (flag == m)printf("YES");
	else printf("NO");
	return 0;
}