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: Runtime Error)

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < n; i++)
#define MA 200000
using namespace std;

long long n, m;
long long d[MA], t[MA];

int main() {
  int bun[MA];
  scanf("%d", &n);
  rep(i, n) {
    scanf("%d", &d[i]);
    bun[d[i] - 1]++;
  }
  scanf("%d", &m);
  rep(i, m) {
    scanf("%d", &t[i]);
  }
  sort(d, d+n);
  bool ans = true;
  rep(i, m) {
    if (bun[t[i] - 1] == 0) {
      ans = false;
      break;
    }
    else {
      bun[t[i] - 1]--;
    }
  }
  cout << (ans ? "YES" : "NO") << '\n';

  return 0;
}