Problem:
Score : 200 points

Problem Statement

You have written N problems to hold programming contests.
The i-th problem will have a score of P_i points if used in a contest.

With these problems, you would like to hold as many contests as possible under the following condition:

A contest has three problems. The first problem has a score not greater than A points, the second has a score between A + 1 and B points (inclusive), and the third has a score not less than B + 1 points.

The same problem should not be used in multiple contests.
At most how many contests can be held?

Constraints

3 \leq N \leq 100

1 \leq P_i \leq 20 (1 \leq i \leq N)

1 \leq A < B < 20

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A B
P_1 P_2 ... P_N

Output

Print the answer.

Sample Input 1

7
5 15
1 10 16 2 7 20 12

Sample Output 1

2

Two contests can be held by putting the first, second, third problems and the fourth, fifth, sixth problems together.

Sample Input 2

8
3 8
5 5 5 10 10 10 15 20

Sample Output 2

0

No contest can be held, because there is no problem with a score of A = 3 or less.

Sample Input 3

3
5 6
5 6 10

Sample Output 3

1

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

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

//int quick(const void *a, const void *b) {
//  return *(int *)a - *(int *)b;
//}

int main(void) {
  int n, p[102], a, b, ans = 0;

  if(scanf("%d %d %d", &n, &a, &b) == 1);

  for(int i = 0; i < n; i++) {
    if(scanf("%d", &p[i]) == 1);
  }

  //qsort(p, sizeof(p) / sizeof(int), sizeof(int), quick);

  for(int i = 0; i < n; i++) {
    if(p[i] >= 1 && p[i] <= a) {
      p[i] = -1;
      for(int j = 0; j < n; j++) {
        if(p[j] > a && p[j] <= b) {
          p[j] = -1;
          for(int k = 0; k < n; k++) {
            if(p[k] > b) {
              ans++;
              p[i] = p[j] = p[k] = - 1;
            }
          }
        }
      }
    }
  }

  printf("%d", ans);


  return 0;
}