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 Fortran (GNU Fortran 9.2.1) (judge verdict: Time Limit Exceeded)

program answer
  implicit none
  integer(8) :: i, j, N, K, ans, f, tmp, b, c
  integer(8), allocatable :: A(:), han(:), se(:)
  read(*,*) N, K
  allocate(A(N),han(N),se(N))
  do i = 1, N
     read(*,*) A(i)
  end do
  ans=1
  do i = 1, N-1
     if(abs(A(i)-A(i+1))<=K) then
        f=i
        exit
     end if
  end do

  ans=1
  han=0
  do i = 1, N-1
     se=0
     if(han(i)/=0) then
        cycle
     end if
     tmp=1
     b=A(i)
     j=1
     do
        if(i+j>=N+1) then
           exit
        end if
        if(han(i+j)/=0) then
           tmp=max(tmp,tmp+han(i+j))
           exit
        else
           if(abs(A(i+j)-b)<=K) then
              se(i)=tmp
              tmp=tmp+1
              b=A(i+j)
              j=j+1
           else
              j=j+1
           end if
        end if
     end do
     han(i)=tmp-se(i)
     ans=max(ans,tmp)
  end do
  
  write(*,*) ans
  stop
 end program answer