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 PyPy3 (7.3.0) (judge verdict: Runtime Error)

n,m=map(int,input().split())
#Union-Find木
#par[i]:iの親　deep[i]:iの深さ　size[i]:iの大きさ
par=[i for i in range(n)]
deep=[1]*n
size=[1]*n

#親を見つける
def find(x):
  if par[x]==x:
    return x
  else:
    return find(par[x])

#二つのグループを統合する
def unite(x,y):
  x=find(x)
  y=find(y)
  if x==y:
    return
  if deep[x]<deep[y]:
    par[x]=y
    size[y]+=size[x]
  else:
    par[y]=x
    size[x]+=size[y]
    if deep[x]==deep[y]:
      deep[x]+=1

#xとyが同じグループに属するかどうか
def same(x,y):
  return find(x)==find(y)

#xが属するグループの要素数を返す
def group_count(x):
  return size[find(x)]

#連結成分の個数を返す
def count():
  check=set()
  for i in par:
    parent=find(i)
    if parent not in check:
      check.add(parent)
  return len(check)

for i in range(m):
  a,b=map(int,input().split())
  unite(a,b)
s=count()
print(s-1)