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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

int main(){
  int n,a,b,num_a,num_b;
  cin >> n;
  int p[n+2];
  cin >> a >> b;
  for (int i=0;i<n;++i){
    cin >> p[i];
  }
  p[n] = a;
  p[n+1] = b;
  sort(p, p+n+2);
  for(int i=n+1;i>0;--i) {
    if(p[i]==a){
      num_a = i;
      continue;
    }
    if(p[i]==b){
      num_b = i;      
      continue;
    }
  }
  cout << min({num_a, num_b-num_a-1, n-num_b+1}) << endl;
  return 0;
}