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>
//<(")
#define ll long long  
using namespace std;

typedef pair<ll, ll> ii;
typedef pair<ll, ii> iii;

const ll MAXX = 1e18;
const ll mod = 1e9 + 7;
const ll mod2 = mod * mod;
const ll siz = 3e5 + 10;
const ll base = 32;

ll n, m;
vector<ll> hold;

ll tree[4 * siz];

void update(ll node, ll left, ll right, ll pos, ll val) {
	if (pos < left || pos > right) {
		return;
	}
	if (left == right) {
		if (pos == left) {
			tree[node] = max(tree[node], val);
		}
		return;
	}
	ll mid = (left + right) / 2;
	update(node * 2, left, mid, pos, val);
	update(node * 2 + 1, mid + 1, right, pos, val);
	tree[node] = max(tree[node * 2], tree[node * 2 + 1]);
}

ll get(ll node, ll left, ll right, ll gLeft, ll gRight) {
	if (gLeft > right || gRight < left) {
		return 0;
	}
	if (gLeft <= left && gRight >= right) {
		return tree[node];
	}
	ll mid = (left + right) / 2;
	return max(get(node * 2, left, mid, gLeft, gRight), get(node * 2 + 1, mid + 1, right, gLeft, gRight));
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	//freopen("INP.txt", "r", stdin);
	//freopen("OUT.txt", "w", stdout);
	cin >> n >> m;
	hold.resize(n + 1);
	ll ans = 0;
	for (ll i = 1; i <= n; i++) {
		cin >> hold[i]; 
	}
	for (ll i = n; i > 0; i--) {
		ll sub = get(1, 1, siz, max(hold[i] - m + 1, 1LL), min(hold[i] + m + 1, siz - 1)) + 1;
		//cerr << sub << '\n';
		ans = max(ans, sub);
		update(1, 1, siz, hold[i] + 1, sub);
	}
	cout << ans << '\n';
	return 0;
}