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>
using namespace std;
const int oo = 0x3f3f3f3f;
const long long ooo = 9223372036854775807ll;
const int _cnt = 1000 * 1000 + 7;
const int _p = 1000 * 1000 * 1000 + 7;
const int N = 200005;
const double PI = acos(-1.0);
const double eps = 1e-9;
int o(int x) { return x % _p; }
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a / gcd(a, b) * b; }
void file_put() {
  string s = "/home/jslijin/jslijin/code/";
  freopen((s + "code.in").c_str(), "r", stdin);
  freopen((s + "code.out").c_str(), "w", stdout);
}
int n, k, a[N], ans;
map<int, int> M;
int main() {
  scanf("%d%d", &n, &k), ans = n;
  for (int i = (1); i <= (n); ++i) scanf("%d", &a[i]), M[a[i]]++;
  sort(a + 1, a + n + 1), n = unique(a + 1, a + n + 1) - a - 1;
  for (int i = (1); i <= (n - 1); ++i)
    ans -= M[a[i]] * (a[i] < a[i + 1] && a[i] + k >= a[i + 1]);
  printf("%d\n", ans);
  return 0;
}
