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)

class SegTree
  attr_accessor(:arr, :size)

	def initialize(n)
		@size = 1<<(n-1).bit_length
		@arr = Array.new(@size*2-1,0)
	end

  def update(i,x)
    i += @size - 1
    @arr[i] = x
    while i > 0
      i = (i-1)/2
      @arr[i] = op(@arr[i*2+1], @arr[i*2+2])
    end
  end

  def query(a,b)
    query_re(a,b+1,0,0,@size)
  end

  def query_re(a,b,k,l,r)
    if r <= a || b <= l
      return -1
    elsif a <= l && r <= b
      @arr[k]
    else
      vl = query_re(a, b, k*2+1, l, (l+r)/2)
      vr = query_re(a, b, k*2+2, (l+r)/2, r)
      op(vl,vr)
    end
  end

  def op(a,b)
    [a,b].max
  end
end

n,k = gets.chomp.split(" ").map(&:to_i)
a = n.times.map{gets.to_i}
seg = SegTree.new(a.max)
n.times do |i|
  l = [a[i]-1-k,0].max
  r = [a[i]-1+k,a.max-1].min
  max = seg.query(l,r)
  seg.update(a[i]-1,max+1)
end
puts seg.query(0,n-1)