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: Runtime Error)

//#include<iostream>
//#include<algorithm>
//#include<string>
//#include <cmath>
//#include <vector>
#include <bits/stdc++.h>

using namespace std;

#define ll long long
#define REP(x,n) for(int x=0;x<n;x++)

//vector//
#define vei vector<int> 
#define velli vector<long long int> 
#define pb(x) push_back(x) // 末尾にxを加える
#define pb2 pop_back() // 末尾削除
#define all(x) (x).begin(),(x).end() //イテレーター
#define be(x) (x).begin()
//auto  auto begin = v.begin()とか

typedef pair<int,int> P;  //q(p.first,p.second)

/// aがbより大きいときaをbにする関数（または小さい ///
//////////////////////////////////////////////////////////////
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
/////////////////////////////////////////////////////////////

int main (){

    // 標準入力
    int n, k; cin >> n >> k;
    vector<int> a(n);
    for(int i=0; i<n; i++){
        cin >> a[i];
    }

     vector<int> dp(n,1); // 0で初期化
    //  dp[0] = 1;

     vector<int> bak(n+3,0);// バケットでその数字の最大値管理 0ind無視
     bak[a[0]] = 1;

     int ans = 0;


    for(int i=1; i<n; i++){
        int maxd = 0;
        if(a[i]-3 >= 1) maxd = max(maxd,bak[a[i]-3]);
        if(a[i]-2 >= 1) maxd = max(maxd,bak[a[i]-2]);
        if(a[i]-1 >= 1) maxd = max(maxd,bak[a[i]-1]);
        maxd = max(maxd,bak[a[i]]);
        maxd = max(maxd,bak[a[i]+1]);
        maxd = max(maxd,bak[a[i]+2]);
        maxd = max(maxd,bak[a[i]+3]);
        dp[i] += maxd;
        chmax(bak[dp[i]],dp[i]);
        chmax(ans,dp[i]);
    }

    cout << ans;
}