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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
 
 
int comp(const void *p, const void *q){
 
    return * (int*) q - * (int*) p;
 
    }
                                                                                       
  
int main(void){
                                                                                                
   
    int n, m, i;
   
    int d[200001];

    int t[200001];

    int a = 0;
   
    int  j = 0;

      
    scanf("%d", &n);

   
    for( i = 0 ; i < n ; i = i + 1 ){

         scanf("%d", &d[i]);

         }

   
    scanf("%d", &m);


    for( i = 0 ; i < m ; i = i + 1 ){

         scanf("%d", &t[i]);

         }


    qsort( d, n, sizeof(int), comp );

    qsort( t, m, sizeof(int), comp );


    for( i = 0 ; i < m ; i = i + 1 ){

         for(  ; j < n ; j = j + 1 ){

               if( t[i] == d[j] ){

               a += 1;

               j = j + 1;

               break;

               }

            }

         }


    if( a == m ) printf("YES\n");

    else printf("NO\n");


    return 0;
                                                          
}