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 Java (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
import java.io.*;
import java.nio.charset.StandardCharsets;

public class Main {

	public static void main(String[] args) throws IOException {
		InputStreamReader reader = new InputStreamReader(System.in, StandardCharsets.UTF_8);
		BufferedReader in = new BufferedReader(reader);
		Main ins = new Main(in);
		ins.calc();
		ins.showResult();
	}

	int N, K;
	int[] A;

	static class Tuple {
		int val, len;

		Tuple(int val, int len) {
			this.val = val;
			this.len = len;
		}
	}

	Main(BufferedReader in) throws IOException {
		String[] tokens = in.readLine().split(" ");
		N = Integer.parseInt(tokens[0]);
		K = Integer.parseInt(tokens[1]);
		A = new int[N];
		for (int i = 0; i < N; ++i) {
			A[i] = Integer.parseInt(in.readLine());
		}
	}

	void calc() {
		Queue<Tuple> que = new PriorityQueue<>(new Comparator<Tuple>() {
			@Override
			public int compare(Tuple o1, Tuple o2) {
				return -1 * Integer.compare(o1.len, o2.len);
			}
		});

		que.add(new Tuple(A[0], 1));
		List<Tuple> list = new ArrayList<>(N);
		for (int i = 1; i < N; ++i) {
			// 追加できるやつを探す
			Tuple tuple = null;
			while (!que.isEmpty()) {
				if (Math.abs(A[i] - que.peek().val) <= K) {
					tuple = que.poll();
					break;
				} else {
					list.add(que.poll());
				}
			}
			if (tuple == null) {
				// 見つけられなかった
				que.addAll(list);
				que.add(new Tuple(A[i], 1));
			} else {
				// 見つけられた
				que.add(new Tuple(A[i], tuple.len + 1));
				que.add(tuple);
				que.addAll(list);
			}
			list.clear();
		}
		System.out.println(que.peek().len);

	}

	void showResult() {
	}
}