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: Time Limit Exceeded)

#include<stdio.h>
int main(void)
{
  int numq,numd,i,j,count=1;
  scanf("%d",&numq);
  int quest[numq];
  for(i=0; i<numq; i++)
  {
    scanf("%d", &quest[i]);
  }
  scanf("%d",&numd);
  int diff[numd];
  for(i=0; i<numd; i++)
  {
    scanf("%d", &diff[i]);
  }
  
  for(i=0; i<numd; i++)
  {
    for(j=0; j<numq; j++)
    {
      if(diff[i]==quest[j])
      {
        quest[j]=0;
        break;
      }
      else if(j==numq-1) count=0;
    }
  }
  if(count==0) printf("NO");
  else printf("YES");
  return 0;
}