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)

#pragma GCC optimize("Ofast", "unroll-loops")

#include <bits/stdc++.h>

using namespace std;

#define ll long long
#define ld long double

template<typename T>
struct SegmentTree {
private:
	int sz, n;
	vector<T> data;
	function<T(T, T)> f;
	T identity_element;
public:
	/* constructor */
	SegmentTree(
		const vector<T>& v, // initial data
		T identity_element, // identity element
		function<T(T, T)> f // operation
	) {
		sz = v.size();
		n = 1; while (n < sz) n <<= 1;
		this->f = f;
		this->identity_element = identity_element;
		data.resize(2 * n - 1, identity_element);
		for (int i = 0; i < sz; ++i)
			data[i + n - 1] = v[i];
		for (int i = n - 2; i >= 0; --i)
			data[i] = f(data[2 * i + 1], data[2 * i + 2]);
	}
	/* update query */
	void update(int idx, const T val) {
		idx += n - 1;
		data[idx] = val;
		while (idx) {
			idx = (idx - 1) >> 1;
			data[idx] = f(data[2 * idx + 1], data[2 * idx + 2]);
		}
	}
	/* get query */
	T get(int left, int right, int k = 0, int l = 0, int r = -1) {
		if (r < 0) r = n;
		if (r <= left || right <= l) return identity_element;
		if (left <= l && r <= right) return data[k];
		T val_l = get(left, right, 2 * k + 1, l, (l + r) / 2);
		T val_r = get(left, right, 2 * k + 2, (l + r) / 2, r);
		return f(val_l, val_r);
	}
};

int main(void){
	int N, K;
	cin >> N >> K;
	SegmentTree<int> dp(
		vector<int>(300001, 0), 0, 
		[](int a, int b){ return max(a, b); }
	);
	while (N--){
		int a; cin >> a;
		int l = max(0, a - K);
		int r = min(300000, a + K) + 1;
		int tmp = dp.get(l, r);
		dp.update(a, tmp + 1);
	}
	int ans = dp.get(0, 300001);
	cout << ans << endl;
	return 0;
}