#include<bits/stdc++.h>
using namespace std;
int n,d;
int a[200005];
map<int,int> mp;
map<int,int> nrm;
int cate;
int aib[400005][2];
int qry(int poz, int care)
{
int aux=0;
for(int i=poz;i>0;i-=i&(-i))
aux = max(aux, aib[i][care]);
return aux;
}
void upd(int poz, int newv, int care)
{
for(int i=poz;i<=cate;i+=i&(-i))
aib[i][care] = max(aib[i][care], newv);
}
int solve_minus()
{
for(int i=1;i<=n;i++)
{
upd(nrm[a[i]],qry(nrm[a[i]]-1,0)+1,0);
upd(nrm[a[i]],qry(nrm[a[i]]-1,1)+1,1);
upd(nrm[a[i]-d],qry(nrm[a[i]],0),1);
}
return max(qry(cate,0),qry(cate,1));
}
signed main()
{
cin>>n>>d;
for(int i=1;i<=n;i++)
{
cin>>a[i];
mp[a[i]]++;
mp[a[i]-d]++;
}
for(auto it:mp)
nrm[it.first]=++cate;
cout<<solve_minus();
return 0;
}
/**
8 10
7 3 5 12 2 7 3 4
dp[i][0] = cea mai lunga secventa care se termina cu valoarea i si pe care inca nu s-a folosit operatia
dp[i][1] = cea mai lunga secventa care se termina cu valoarea i si pe care s-a folosit deja operatia
dp[a[i] - d][1] max= dp[orice < a[i]][0] + 1
*/
