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 with AC Library v1.1) (judge verdict: Wrong Answer)

#include <atcoder/all>
#include <bits/stdc++.h>
using namespace std;
using namespace atcoder;
#define lli long long int
#define llu unsigned long long int
#define sa(a,n,t) for(int ii=0;ii<n;ii++) scanf("%"#t"",&(a)[ii])
#define rep(i,n) for(int i=0;i<n;i++) 
#define reps(i,a,n) for(int i=a;i<n;i++) 
#define rrep(i,n) for(int i=n-1;i>=0;i--)
#define rreps(i,a,n) for(int i=n-1;i>=a;i--)
#define validl(i,a,b) ((i>=a)&&(i<=b))?1:0
#define validmat(x,y,a,b,c,d) (validl(x,a,b) && validl(y,a,b))?1:0
#define validmn(x,y,m,n) validmat(x,y,0,m,0,n)
#define valid(i,n) validl(i,0,n)
#define v(t) vector<t>
#define vv(t) vector<vector<t> >
#define mk make_pair
#define pb push_back
#define pa(a,n,t)  for(int i=0;i<n;i++) printf("%"#t"",a[i])
#define p2 pair<int,int>
#define p3 pair<p2,int>
#define fi first
#define se second
#define sd(a) scanf("%d",&a)
#define sl(a) scanf("%lld",&a)
#define pl(a) printf("%lld",a)
#define pd(a) printf("%d",a)
#define sf(a) scanf("%lf",&a)
#define pf(a) printf("%lf",a)
#define nline printf("\n")
#define ss(a) scanf("%s",a)
#define ps(a) printf("%s",a)
#define sc(a) scanf("%c",&a)
#define pc(a) printf("%c",&a)
#define mf(m,f) m.find(f)!=m.end()
#define pp3(m) cout<<m.fi.fi<<" "<<m.fi.se<<" "<<m.se<<" "
#define pp2(m) cout<<m.fi<<" "<<m.se<<" "
#define debug 0

int main() {
    int n,k;
    cin>>n>>k;
    int a[300001]={0};
    int b[300006]={0};
    rep(i,n) cin>>a[i];
    rep(i,n){
        int tmp=1;
        for(int j=3;j>=-3;j--){
            if(a[i]-j>=0)
            tmp=max(tmp,b[a[i]-j]+1);
        }
        b[a[i]]=max(b[a[i]],tmp);
        //cout<<b[a[i]]<<" ";
    }
    cout<<b[a[n-1]]<<endl;
    return 0;
}