CodeForces 990B

You have a Petri dish with bacteria and you are preparing to dive into the harsh micro-world. But, unfortunately, you don't have any microscope nearby, so you can't watch them.

You know that you have $\mathrm{nn bacteria\; in\; the\; Petri\; dish\; and\; size\; of\; the ii-th\; bacteria\; is aiai.\; Also\; you\; know\; intergalactic\; positive\; integer\; constant KK.}$

The $\mathrm{ii-th\; bacteria\; can\; swallow\; the jj-th\; bacteria\; if\; and\; only\; if ai>ajai>aj and ai\le aj+Kai\le aj+K.\; The jj-th\; bacteria\; disappear,\; but\; the ii-th\; bacteria\; doesn\text{'}t\; change\; its\; size.\; The\; bacteria\; can\; perform\; multiple\; swallows.\; On\; each\; swallow\; operation\; any\; bacteria ii can\; swallow\; any\; bacteria jj if ai>ajai>aj and ai\le aj+Kai\le aj+K.\; The\; swallow\; operations\; go\; one\; after\; another.}$

For example, the sequence of bacteria sizes 

Since you don't have a microscope, you can only guess, what the minimal possible number of bacteria can remain in your Petri dish when you finally will find any microscope.

Input

The first line contains two space separated positive integers $\mathrm{nn and KK (1\le n\le 2\cdot 1051\le n\le 2\cdot 105, 1\le K\le 1061\le K\le 106)\; —\; number\; of\; bacteria\; and\; intergalactic\; constant KK.}$

The second line contains $\mathrm{nn space\; separated\; integers a1,a2,\dots ,ana1,a2,\dots ,an (1\le ai\le 1061\le ai\le 106)\; —\; sizes\; of\; bacteria\; you\; have.}$

Output

Print the only integer — minimal possible number of bacteria can remain.

Sample Input

Input

7 1
101 53 42 102 101 55 54

Output

3

Input

6 5
20 15 10 15 20 25

Output

1

Input

7 1000000
1 1 1 1 1 1 1

Output

7

Hint

The first example is clarified in the problem statement.

In the second example an optimal possible sequence of swallows is: 

In the third example no bacteria can swallow any other bacteria.

题意：满足题中所给条件的病菌会发生吞并，问最后剩下的病菌的最少的数量。

吞并的条件有两个，可以通过在先确定一个条件的前提下，判断另一个条件。

先对病菌从大到小排一遍序，然后从后向前遍历一遍，只需要判断相邻的两个是否满足条件即可（特殊的是，有相等的数的时候），因为相邻的两个是差最小的，最容易满足第二个条件。

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<stack>
#include<deque>
#include<map>
#include<iostream>
using namespace std;
typedef long long  LL;
const double pi=acos(-1.0);
const double e=exp(1);
const int N = 200010;

#define lson i << 1,l,m
#define rson i << 1 | 1,m + 1,r

map<LL,LL> mp;
LL con[N];

bool cmp(LL a,LL b)
{
    return a>b;
}
int main()
{
    LL i,p,j,x;
    LL n,k;
    scanf("%lld%lld",&n,&k);
    for(i=0;i<n;i++)
    {
        scanf("%lld",&con[i]);
        mp[con[i]]++;
    }
    sort(con,con+n,cmp);
    p=n;
    for(i=n-2;i>=0;i--)
    {
        if(con[i]>con[i+1]&&con[i]<=(con[i+1]+k))
        {
            p=p-mp[con[i+1]];
        }
    }
    printf("%lld
",p);
    return 0;
}