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(){
  int n1, n2, d[2][200000] = {0}, a, k = 0, j;
  scanf("%d", &n1);
  for(int i = 0; i < n1; ++i){
    scanf("%d", &d[1][i]);
    d[0][i] = 1;
  }
   for (int i = 0; i < n1 - 1; i++){
    for (j = n1 - 1; j >= i + 1; j--){   //　右から左に操作
      if (d[1][j] < d[1][j-1]) { 
        int tmp = d[1][j];
        d[1][j] = d[1][j-1];
        d[1][j-1] = tmp;
      }
    }
  }
  scanf("%d", &n2);
  for(int i = 0; i < n2; ++i){
    int l = 0;
    scanf("%d", &a);
    if(k) continue;
    if(a > d[1][n1 - 1]){
      k++;
      continue;
    }
    for(j = 0; j < n1; ++j){
      while(1){
        if(j + 50000 > n1) break;
        if(d[1][j + 50000] < a){
          j += 50001;
          continue;
        }
        break;
      }
      if(d[1][j] > a){
        k++;
        break;
      }
      if(d[0][j] == 0) continue;
      if(d[1][j] == a){
        d[0][j] = 0;
        l = 1;
        break;
      }
    }
    if(j == n1 - 1 && l == 0) k++;
  }
  if(k) puts("NO");
  else puts("YES");
  return 0;
}