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 <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef long long int value_type; /* ソートするキーの型 */
typedef long long int lli;

value_type med3(value_type x, value_type y, value_type z)
/* x, y, z の中間値を返す */
{
  if (x < y)
    if (y < z)
      return y;
    else if (z < x)
      return x;
    else
      return z;
  else if (z < y)
    return y;
  else if (x < z)
    return x;
  else
    return z;
}

void quicksort(value_type a[], int left, int right)
/* クイックソート
 * a     : ソートする配列
 * left  : ソートするデータの開始位置
 * right : ソートするデータの終了位置
 */
{
  if (left < right) {
    int i = left, j = right;
    value_type tmp,
        pivot = med3(a[i], a[i + (j - i) / 2],
                     a[j]); /* (i+j)/2ではオーバーフローしてしまう */
    while (1) { /* a[] を pivot 以上と以下の集まりに分割する */
      while (a[i] < pivot) i++; /* a[i] >= pivot となる位置を検索 */
      while (pivot < a[j]) j--; /* a[j] <= pivot となる位置を検索 */
      if (i >= j) break;
      tmp = a[i];
      a[i] = a[j];
      a[j] = tmp; /* a[i],a[j] を交換 */
      i++;
      j--;
    }
    quicksort(a, left, i - 1);  /* 分割した左を再帰的にソート */
    quicksort(a, j + 1, right); /* 分割した右を再帰的にソート */
  }
}

int main() {
  int i, j;
  lli* d = (lli*)malloc(sizeof(lli) * 200001);
  int* dnum = (int*)malloc(sizeof(int) * 200001);
  lli* t = (lli*)malloc(sizeof(lli) * 200001);
  int* tnum = (int*)malloc(sizeof(int) * 200001);

  int N, M;
  lli temp;
  scanf("%d", &N);
  for (i = 0; i < N; i++) {
    scanf("%lld", &d[i]);
  }
  scanf("%d", &M);
  /*
  if(M>N) {
    printf("No");
  }
  */
  for (i = 0; i < M; i++) {
    scanf("%lld", &t[i]);
  }

  quicksort(d, 0, N - 1);
  quicksort(t, 0, M - 1);
  /*
  printf("N: \n");
  for (i = 0; i < N; i++) {
    printf("%lld\n", d[i]);
  }
  printf("M; \n");
  for (i = 0; i < M; i++) {
    printf("%lld\n", t[i]);
  }
  */
  int flag = 0;
  int nd_flag = 0;
  int cur = 0;
  for (i = 0; i < M; i++) {
    //printf("Go\n");
    flag = 0;
    for (j = cur; j < N; j++) {
      if (t[i] == d[j]) {
        cur = j + 1;
        d[j] = 0;
        flag = 1;
        //printf("t: %d, d: %d\n", t[i], d[j]);
        break;
      } else if (d[j] > t[i]) {
        break;
      }
    }
    if (flag == 0) {
      nd_flag = 1;
      break;
    }
  }
  /*
    for (i = 0; i < M; i++) {
      flag = 0;
      for (j = 0; j < N; j++) {
        if (t[i] == d[j]) {
          flag = 1;
          d[j] = 0;
          break;
        }
      }
      if (flag == 0) {
        nd_flag = 1;
        break;
      }
    }
    */

  if (nd_flag != 1) {
    printf("YES");
  } else {
    printf("NO");
  }
  return 0;
}