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: Accepted)

#include <bits/stdc++.h>
//#include <atcoder/all>
using namespace std;
//using namespace atcoder;
#define rep(i,n) for(int i = 0; i < (n);i++)
#define sz(x) int(x.size())
typedef long long ll;
typedef long double ld;
typedef pair<int,int> P;

template <typename T> 
struct SegmentTree {
  
  int n;
  vector<T> node;
  T e;
 
  T f(T a, T b) {return max(a, b);}
  void g(T &a, T b) {a = b;}
 
  SegmentTree (int sz, T init = 0) {
    n = 1; e = init; while (n < sz) n <<= 1;
    node.assign(2*n, init);
    for (int i = n - 2; i >= 0; i--) node[i] = f(node[2*i+1], node[2*i+2]);
  }
 
  void update(int k, T x) {
    k += (n - 1);
    g(node[k], x);
    while (k > 0) {
      k = (k - 1) >> 1;
      node[k] = f(node[2*k+1], node[2*k+2]);
    } 
  }
 
  T query(int a, int b, int k = 0, int l = 0, int r = -1) {
    if (r < 0) r = n;
    if (r <= a || b <= l) return e;
    if (a <= l && r <= b) return node[k];
    T vl = query(a, b, 2*k+1, l, (l+r)/2);
    T vr = query(a, b, 2*k+2, (l+r)/2, r);
    return f(vl, vr);
  }
 
  T getvalue(int k) {
    return query(k,k+1);
  }
  
  template <typename F> 
  int find (int a, int b, T x, F &ok, int k = 0, int l = 0, int r = -1) {
    if (r < 0) r = n;
    if (!ok(node[k]) || r <= a || b <= l) return b;
    if (k >= n - 1) return k - n + 1;
    int vl = find(a, b, x, ok, 2*k+1, l, (l + r)/2);
    if (vl != b) return vl;
    return find(a, b, x, ok, 2*k+2, (l + r)/2, r);
  }
 
};
int main() {
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  rep(i,n) cin >> a[i];
  SegmentTree<int> dp(300010);

  for (int i = 0; i < n; i++) {
    int l = max(0, a[i] - k), r = min(300000, a[i] + k);
    int b = dp.query(l, r + 1) + 1;
    if (b > dp.query(a[i], a[i] + 1)) dp.update(a[i], b);
  }

  cout << dp.query(0, 300010) << endl;
  return 0; 
}