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>

void merge(long s[], int low, int middle, int high)
{
 int i;
 int j;
 long arr1[200005];
 long arr2[200005];
 int ind1=0;
 int ind2=0;

 for (i=low;i<=middle; i++) arr1[i-low]=s[i];
 for (i=middle+1; i<=high; i++) arr2[i-middle-1]=s[i];

 i=low;
 
 for(;;)
 {  
   if(arr1[ind1]<=arr2[ind2])
   {
     s[i++]=arr1[ind1++];
   }else{
     s[i++]=arr2[ind2++];
   }
   if(ind1>middle-low) break;
   if(ind2>high-middle-1) break;
 }
 for(j=ind1;j<=middle-low;j++) s[i++]=arr1[j];
 for(j=ind2;j<=high-middle-1;j++) s[i++]=arr2[j];

}


void mergesort(long s[], int low, int high)
{
 int middle;
 if(low<high)
 {
  middle=(low+high)/2;
  mergesort(s,low,middle);
  mergesort(s,middle+1,high);
  merge(s,low,middle,high);
 }
}

int main()
{
 int N;
 long D[200005];
 int M;
 long T[200005];
 int i;
 int start=0;
 int allfound=1;
 int found[200005];
 int j;

 scanf("%d",&N);
 for(i=0;i<N;i++) scanf("%ld",&D[i]);
 scanf("%d",&M);
 for(i=0;i<M;i++) scanf("%ld",&T[i]);
 for(i=0;i<M;i++) found[i]=0;

 mergesort(D,0,N-1);
 mergesort(T,0,M-1 );

 for(i=0;i<N;i++)
 {
  for(j=start;j<M;j++)
  {
   if(D[i]==T[j])
   {
    start=j+1;
    found[j]=1;
    break;
   }
   if(D[i]>T[j]) break;
  }
 }

 for(i=0;i<M;i++) 
 {
  if(found[i]!=1) 
  {
   allfound=0;  
   break;
  }
 }

 if(allfound==1)
 {
  printf("YES\n");
 }else{
  printf("NO\n");
 }

 return 0;
}