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>

int cmpnum(const void * n1, const void * n2)
{
	if (*(long *)n1 > *(long *)n2)
	{
		return 1;
	}
	else if (*(long *)n1 < *(long *)n2)
	{
		return -1;
	}
	else
	{
		return 0;
	}
}

int main(void) {

  long n;
  scanf("%ld", &n);
  long d[n];
  for (long i = 0; i < n; i++) {
    scanf("%ld", &d[i]);
  }
  long m;
  scanf("%ld", &m);
  long t[m];
  for (long i = 0; i < m; i++) {
    scanf("%ld", &t[i]);
  }
  qsort(d, n, sizeof(long), cmpnum);
  qsort(t, m, sizeof(long), cmpnum);
  int flag;
  long idea = 0;
  for (long i = 0; i < m; i++) {
    if (idea >= n) {
      flag = 0;
      break;
    }
    flag = 1;
    while (1) {
      if (idea >= n) {
        flag = 0;
        break;
      }
      if (d[idea] == t[i]) {
        idea++;
        break;
      } else if (d[idea] < t[i]) {
        idea++;
      } else {
        flag = 0;
        break;
      }
    }
    if (flag == 0) {
      break;
    }
  }
  if (flag == 1) {
    printf("YES\n");
  } else {
    printf("NO\n");
  }

  return 0;
}