Problem:
Score : 400 points

Problem Statement

You are given a sequence A_1, A_2, ..., A_N and an integer K.

Print the maximum possible length of a sequence B that satisfies the following conditions:

B is a (not necessarily continuous) subsequence of A.

For each pair of adjacents elements of B, the absolute difference of the elements is at most K.

Constraints

1 \leq N \leq 300,000

0 \leq A_i \leq 300,000

0 \leq K \leq 300,000

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
A_1
A_2
:
A_N

Output

Print the answer.

Sample Input 1

10 3
1
5
4
3
8
6
9
7
2
4

Sample Output 1

7

For example, B = (1, 4, 3, 6, 9, 7, 4) satisfies the conditions.

It is a subsequence of A = (1, 5, 4, 3, 8, 6, 9, 7, 2, 4).

All of the absolute differences between two adjacent elements (|1-4|, |4-3|, |3-6|, |6-9|, |9-7|, |7-4|) are at most K = 3.

Solution in Ruby (2.7.1) (judge verdict: Time Limit Exceeded)


### SNIPPET
  # n = gets.split.map(&:to_i)
  # array = n.times.map { gets.split.map(&:to_i) }
  # [].all?(&:even?)
  # a = [*1..m].repeated_combination(n).to_a
  # [1,2,3,4,5].select { |num| num.even? }  # => [2, 4]
  # ["a","a","b"].group_by(&:itself).map{|k,v| [k, v.count]}.to_h #=> {"a"=>2, "b"=>1}
  # 切り捨て: .floor(2).to_f ,切り上げ： .ceil(2).to_f,四捨五入: round(2)
  # 3.upto(6) do |i|, 6.downto(3) do |i|
  # 公約数125.gcd(100)、公倍数125.lcm(100)
  # PI = Math::PI
  # 高さ = a * Math.sin(w / 180.0 * Math::PI), 底辺 = a * Math.cos(w / 180.0 * Math::PI)
  def chmax(a, b) a > b ? a : b end
  # INF = Float::INFINITY
  # def chmin(a, b) a < b ? a : b end

# require 'pry'
n,K = gets.split.map(&:to_i)

array = n.times.map { gets.to_i } 
max = 0
(1<<n).times do |x|
  ary = []
  n.times do |i|
    if x[i] == 1
      ary.push(array[i])
    end
  end
  next if ary.length <= max
  flag = true
  0.upto(ary.length - 1 - 1) do |k|
    # binding.pry
    unless K >= (ary[k] - ary[k+1]).abs
      flag = false
    end
  end
  max = chmax(max, ary.length) if flag 
end

puts max