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: Wrong Answer)

N, K = gets.split.map(&:to_i)
ary = []
N.times do 
    ary << gets.to_i
end
max_cnt = 0
hash = {}
hash[N-1] = 1
(N-2).downto(0) do |n|
    cnt = 1
    target_idx = n
    n.upto(N-2) do |m|
        if (ary[target_idx] - ary[m+1]).abs <= K
            if hash[m+1]
                cnt += hash[m+1]
                break
            else
                cnt += 1
                target_idx = m+1
            end
        end
    end
    max_cnt = [max_cnt, cnt].max
    hash[n] = cnt
end
puts max_cnt