2 1
1 1
 4 1
2 2 1 1
 10 1
2 6 3 4 2 4 4 3 2 1
 2 1
999152 999153
 8 1000000
1 1 5 1000000 1000000 2 2 2
 9 2
1 6 1 5 5 8 6 8 7
 15 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
 1 4
8
 10 1
6 3 1 3 6 4 1 3 6 4
 10 1
1 2 3 5 6 8 10 11 9 4
 7 2
1 5 5 8 9 8 8
 2 1000000
1 1000000
 8 2
3 13 9 8 3 13 9 14
 1 1
1
 2 1
1 2

2
 2
 4
 1
 2
 4
 15
 1
 7
 2
 4
 1
 5
 1
 1

#include <bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
// #include <boost/multiprecision/cpp_int.hpp> 

#define gc getchar//_unlocked
#define pc putchar//_unlocked
#define ll long long
#define ld long double
#define pb push_back
#define mp make_pair
#define pp pair<int,int>
#define ppl pair<ll,ll>
#define bigint boost::multiprecision::cpp_int
#define finp ios_base::sync_with_stdio(0);cin.tie(0);
#define bc __builtin_popcountll
#define afor(i,a,b) for(int i=a;i<=b;++i)
#define bfor(i,a,b) for(int i=a;i>=b;--i)
#define vi vector<int>
#define vpp vector<pp>
#define vll vector<ll>
#define fr first
#define se second

using namespace std;
using namespace __gnu_pbds;

char putnb[20];
void putn(ll n) {if(!n)pc('0');if(n<0)pc('-'),n=0-n;int pi=0;while(n)putnb[pi++]=(n%10)+'0',n/=10;while(pi)pc(putnb[--pi]);}
void sci(int *x) {register char c = gc();*x = 0;for(; (c<48)||(c>57);c = gc());for(; (c>47)&&(c<58);c = gc())*x = (int)((((*x)<<1) + ((*x)<<3)) + c - 48);}
void scll(ll *x)  {register char c = gc();*x = 0;for(; (c<48)||(c>57);c = gc());for(; (c>47)&&(c<58);c = gc())*x =  (ll)((((*x)<<1) + ((*x)<<3)) + c - 48);}
ll fp(ll a,ll b,ll c) {if(b==0)return 1%c; if(b==1)return a%c; ll ret=fp(a,b/2,c); ret=(ret*ret)%c; if(b&1)ret=(ret*a)%c; return ret;}

const ll mod=1e9 +7;
const ll mod2=1999999973;
const ll inf=1e18;
const int infs=1e9 + 1000;
const int N=200000;
const long double PI = acos(-1);

template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;


ll arr[N + 5];
int main()
{
  finp;

  ll n,k;
  cin>>n>>k;
  afor(i,1,n)cin>>arr[i];
  sort(arr+1,arr+n+1);
  ll ans = 1;
  afor(i,1,n-1)
  {
    if(arr[i+1]>(arr[i]+k) || arr[n]==arr[i])++ans;
  }
  cout<<ans;
return 0;
}