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: Runtime Error)

class RMQ
  def initialize(ini)
    @a = []
    @v = Array.new(ini + 1, 0)
    i = 1
    while i < ini
      @a << Array.new(i, 0)
      i *= 2
    end
    @a << @v
  end

  def larger(x, y)
    x > y ? x : y
  end

  def inspect
    @a.inject(""){|acc, x| acc + x.join(' ') + "\n"}
  end

  def []=(i, x)
    @v[i] = x
    layer = @a.size - 2
    while layer >= 0
      i /= 2
      y = larger(@a[layer + 1][i * 2], @a[layer + 1][i * 2 + 1])
      break if @a[layer][i] > y
      @a[layer][i] = y
      layer -= 1
    end
  end

  def rangeMax(a, b, layer = @a.size - 1)
    if a > b
      return 0
    else
      side = larger((a.odd? ? @a[layer][a] : 0), (b.even? ? @a[layer][b] : 0))
      larger(side, rangeMax((a + 1) / 2, (b - 1) / 2, layer - 1))
    end
  end
end
 
tree = RMQ.new(300_000)
 
N, K = gets.split.map(&:to_i)
N.times do
  a = gets.to_i
  left = a - K
  left = 0 if left < 0
  right = a + K
  right = 300_000 if right > 300_000
  tree[a] = 1 + tree.rangeMax(left, right)
end
puts tree.rangeMax(0, 300_000)