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 <bits/stdc++.h>
#define REP(i,n) for (int i = 0; i <(n); ++i)
#define REP2(i,x,n) for (int i = x; i <(n); ++i)
#define ALL(v) v.begin(), v.end()
#define RALL(v) v.rbegin(), v.rend()
using namespace std;
using ll = long long;
using P = pair<int,int>;
static const double PI = acos(-1);
static const int INF = 1e9+7;
//debug 
#ifdef _DEBUG
#define debug(var)  do{cout << #var << " :";view(var);}while(0)
#else
#define debug(...)
#endif
template<typename T> void view(T e){cout << e << endl;}
template<typename T> void view(const vector<T>& v){for(const auto& e : v){ cout << e << " "; } cout << endl;}
template<typename T> void view(const vector<vector<T> >& vv){ for(const auto& v : vv){ view(v); } }


ll seg[1 << 20];

void set_value(ll pos, ll val){
  pos += 1 << 19; //20?
  seg[pos] = val;
  while((pos /= 2) > 0 ){
    seg[pos] = max(seg[pos * 2], seg[pos * 2 + 1]);//自分の左右の子供のminをとって更新
  }
}

//クエリの右橋，左端，今いる場所の区間の左右端，今いる場所 デフォルト引数設定しとくとよい
ll get_max(ll ql, ll qr, ll sl = 0, ll sr = 1 << 19, ll pos = 1){
  //全くかぶっていないとき
  if(qr <= sl || sr <= ql) return 0;  //意味のない数字を返す　maxのときは，0を返す
  //完全にかぶっているとき（セグメントがクエリに）
  if(ql <= sl && sr <= qr) return seg[pos];
  //
  ll sm = (sl + sr) / 2;
  ll lmax = get_max(ql,qr,sl,sm,pos * 2); //左右の子に託す
  ll rmax = get_max(ql,qr,sm,sr,pos * 2 + 1); //左右の子に託す
  return max(lmax, rmax);
}


int main(){
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  REP(i,n) cin >> a[i];
  REP(i,300010){
    set_value(i,0);
  }

  REP(i,n){
    int add = get_max(a[i]-k, a[i]+k);
    set_value(a[i], add+1);
  }



  int ans = get_max(0,300010);
  cout << ans+1 << endl;
  return 0;
}