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 C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

//include
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <cmath> 
#include <iomanip>
#include <math.h>
#include <utility>
#include <functional>
#include <cstdlib>
#include <map>
#include <utility>
//using
using namespace std;
using vl = vector <long long>;
using vs = vector <string>;
using vc = vector <char>;
using ll= long long;
using vvl = vector<vector<ll> >;
using vvc = vector<vector<char> >;
//vector<vector<char> > hyou(N, vector<char>(N));
using vd = vector <double>;
//define
//#define int long long
#define rep(i,n) for(int i=0; i<n; i++)
#define print(n) cout<<n<<endl;
#define sortp(d) sort(d.begin(),d.end()) //1234
#define sortm(d) sort(d.rbegin(),d.rend()) //4321
//素数判定　O(√A) 
bool is_prime(int x){
 if(x<=1) return false;
 for(int i=2;i*i<=x;i++)
 if(x%i==0) return false;
 return true;
 }
 //各位の和　
int wa(ll x)
{
ll sum = 0;
while(x!=0){
  sum += x% 10;
  x /= 10;
}
return sum;
}
/*順列生成
do {
  // 順列に対する処理
} while (next_permutation(配列変数.begin(), 配列変数.end()));*/

//絶対値　abs()
//文字→数字　'0' アルファベット→数字　'a'　'A'が65番,'a'が97番だわ
//型キャスト　char('a'+1)
//reverse 回文
//if(isupper('A'))->大なら真


//グローバル変数宣言
ll p=0,q=0,r=0;

int main()
{
  ll n,k;
  cin>>n>>k;
  vl a(n);
  rep(i,n)
  {
    cin>>a[i];
  }
  q=a[0];
  for (int i = 1; i < n ;i++)
  {
    if (abs(a[i]-q)<=k)
    {
      p++;
      q=a[i];
    }
    
  }
  print(p+1);
}