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: Wrong Answer)

#include <stdio.h>
#define MAX 1000000000

int main(){ 

	int i, j;
    int M, N;

    scanf("%d", &N); // 問題候補数


    int n_i[N];

    // printf("%d", n_number);

    scanf("%d", &n_i[0]);

    char temp;

    for(i=1; i<N; i++){
        scanf("%c", &temp);
        scanf("%d", &n_i[i]);
    }

    /*
    for(i=0;i<N; i++){
        printf("%d\n", n_i[i]);
    }
    */


    scanf("%d", &M);
    int m_i[M];

    scanf("%d", &m_i[0]);

    for(i=1; i<M; i++){
        scanf("%c", &temp);
        scanf("%d", &m_i[i]);
    }

    /*
    printf("%d, %d\n", N, M);
    */

    for (i=0; i<M; i++) {
        j=0;

        while(j<N){
            if(m_i[i] == n_i[j]) {
                    m_i[i] = n_i[j] = 0;
                    break;    
            }        
            ++j;
        }    
    }


    /*
   for(i=0; i<M; i++){
       printf("%d\n", m_i[i]);
    }
    */

    int check = 1;

    for(i=0; i<M; i++){
        check *= m_i[i];
    }

    if(check == 1){
        printf("YES");
    }else{
        printf("NO");
    }



return 0;

}