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: Time Limit Exceeded)

#include <iostream>
using namespace std;

int test(int n, int a, int b, int p[]);

int main() {
	int N, A, B;	int ans = 0;
	cin >> N;
	cin >> A >> B;
	int P[N];
	for (int i = 0; i < N; i++) {
		cin >> P[i];
	}

	ans = test(N, A, B, P);

	cout << ans << endl;

	return 0;
}

int test(int n, int a, int b, int p[]) {
	int ans = 0;	int t = 0;
	int sum[] = { 0 };
	for (int i = 0; i < n; i++) {
		sum[i] += 1;
	}

	while (1) {
		t = 0;
		for (int i = 0; i < n; i++) {
			if ((sum[i] == 1) || (p[i] <= a)) {
				t += 1;
				sum[i] = 0;
			}
		}
		if (t == 0) break;

		for (int i = 0; i < n; i++) {
			if ((sum[i] == 1) || (p[i] > a) || (p[i] <= b)){
				t += 1;
				sum[i] = 0;
			}
		}
		if (t == 1) break;

		for (int i = 0; i < n; i++) {
			if ((sum[i] == 1) || (p[i] > b)){
				t += 1;
			sum[i] = 0;
			}
		}
		if (t == 2) break;
		ans += 1;
	}

	return (ans);
}