You are policeman and you are playing a game with Slavik. The game is turn-based and each turn consists of two phases. During the first phase you make your move and during the second phase Slavik makes his move.

There are n doors, the i-th door initially has durability equal to a_i.

During your move you can try to break one of the doors. If you choose door i and its current durability is b_i then you reduce its durability to max(0, b_i - x) (the value x is given).

During Slavik's move he tries to repair one of the doors. If he chooses door i and its current durability is b_i then he increases its durability to b_i + y (the value y is given). Slavik cannot repair doors with current durability equal to 0.

The game lasts 10^{100} turns. If some player cannot make his move then he has to skip it.

Your goal is to maximize the number of doors with durability equal to 0 at the end of the game. You can assume that Slavik wants to minimize the number of such doors. What is the number of such doors in the end if you both play optimally?

Input

The first line of the input contains three integers n, x and y (1 ≤ n ≤ 100, 1 ≤ x, y ≤ 10^5) — the number of doors, value x and value y, respectively.

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5), where a_i is the initial durability of the i-th door.

Output

Print one integer — the number of doors with durability equal to 0 at the end of the game, if you and Slavik both play optimally.

Examples

Input


6 3 2
2 3 1 3 4 2


Output


6


Input


5 3 3
1 2 4 2 3


Output


2


Input


5 5 6
1 2 6 10 3


Output


2

Note

Clarifications about the optimal strategy will be ignored.
#!/usr/bin/env python
import os
import sys

from io import BytesIO, IOBase

# Bootstrap https://github.com/cheran-senthil/PyRival/blob/master/tests/misc/test_bootstrap.py
from types import GeneratorType


def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to

    return wrappedfunc


#################

from collections import defaultdict
from math import sqrt

INF = pow(2, 63)


def isPrime(number):
    """
    input: positive integer 'number'
    returns true if 'number' is prime otherwise false.
    """

    # precondition
    assert isinstance(number, int) and (
            number >= 0
    ), "'number' must been an int and positive"

    status = True

    # 0 and 1 are none primes.
    if number <= 1:
        status = False

    for divisor in range(2, int(round(sqrt(number))) + 1):

        # if 'number' divisible by 'divisor' then sets 'status'
        # of false and break up the loop.
        if number % divisor == 0:
            status = False
            break

    # precondition
    assert isinstance(status, bool), "'status' must been from type bool"

    return status


def prime_factors(n: int):
    """
    Returns prime factors of n as a list.
    >>> prime_factors(0)
    []
    >>> prime_factors(100)
    [2, 2, 5, 5]
    >>> prime_factors(2560)
    [2, 2, 2, 2, 2, 2, 2, 2, 2, 5]
    >>> prime_factors(10**-2)
    []
    >>> prime_factors(0.02)
    []
    >>> x = prime_factors(10**241) # doctest: +NORMALIZE_WHITESPACE
    >>> x == [2]*241 + [5]*241
    True
    >>> prime_factors(10**-354)
    []
    >>> prime_factors('hello')
    Traceback (most recent call last):
        ...
    TypeError: '<=' not supported between instances of 'int' and 'str'
    >>> prime_factors([1,2,'hello'])
    Traceback (most recent call last):
        ...
    TypeError: '<=' not supported between instances of 'int' and 'list'
    """
    i = 2
    factors = []
    while i * i <= n:
        if n % i:
            i += 1
        else:
            n //= i
            factors.append(i)
    if n > 1:
        factors.append(n)
    return factors


def greatest_common_divisor(a: int, b: int) -> int:
    """
    Calculate Greatest Common Divisor (GCD).
    >>> greatest_common_divisor(24, 40)
    8
    >>> greatest_common_divisor(1, 1)
    1
    >>> greatest_common_divisor(1, 800)
    1
    >>> greatest_common_divisor(11, 37)
    1
    >>> greatest_common_divisor(3, 5)
    1
    >>> greatest_common_divisor(16, 4)
    4
    >>> greatest_common_divisor(-3, 9)
    3
    >>> greatest_common_divisor(9, -3)
    3
    >>> greatest_common_divisor(3, -9)
    3
    >>> greatest_common_divisor(-3, -9)
    3
    """
    return abs(b) if a == 0 else greatest_common_divisor(b % a, a)


def solveb(n, tuparr):
    return -1


import math


def solve(n, x, y, arr):
    if x > y:
        print(n)
        return
    c = sum([1 for d in arr if y - x <= d <= x])
    c1 = sum([1 for d in arr if d < y - x])
    print(c1 + c // 2 + c % 2)


def main():
    # n, k = map(int, input().strip().split())
    # a = []
    # for _ in range(n):
    #     a.append(input().strip())
    # print(solve(n, k, a))
    # for _ in range(int(input().strip())):
    # n = int(input().strip())
    [n, m, o] = list(map(int, input().strip().split()))
    arrn = list(map(int, input().strip().split()))
    # sarr = []
    # for _ in range(n):
    #     sarr.append(input().strip())

    solve(n, m, o, arrn)


def mainb():
    pass


######## Python 2 and 3 footer by Pajenegod and c1729

# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.

# So on cf, use PyPy2 for best string performance.

py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip

    range = xrange

import os, sys
from io import IOBase, BytesIO

BUFSIZE = 8192


class FastIO(BytesIO):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0, 2), super(FastIO, self).write(s))[0])
        return s

    def read(self):
        while self._fill(): pass
        return super(FastIO, self).read()

    def readline(self):
        while self.newlines == 0:
            s = self._fill();
            self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s: self.buffer.write(s.encode('ascii'))
            self.read = lambda: self.buffer.read().decode('ascii')
            self.readline = lambda: self.buffer.readline().decode('ascii')


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline()

# Cout implemented in Python
import sys


class ostream:
    def __lshift__(self, a):
        sys.stdout.write(str(a))
        return self


cout = ostream()
endl = '\n'


# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero=0):
    conv = ord if py2 else lambda x: x
    A = [];
    numb = zero;
    sign = 1;
    i = 0;
    s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'0'[0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-'[0]:
                sign = -1
            elif s[i] != b'\r'[0]:
                A.append(sign * numb)
                numb = zero;
                sign = 1
            i += 1
    except:
        pass
    if s and s[-1] >= b'0'[0]:
        A.append(sign * numb)
    return A


if __name__ == "__main__":
    #for _ in range(3):  # todo for testing else remove for loop
        main()
