990B - Micro-World

B. Micro-World

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 $n$ bacteria in the Petri dish and size of the $i$-th bacteria is $a_i$. Also you know intergalactic positive integer constant $K$.

The $i$-th bacteria can swallow the $j$-th bacteria if and only if $a_i>a_j$ and $a_i\le a_j+K$. The $j$-th bacteria disappear, but the $i$-th bacteria doesn't change its size. The bacteria can perform multiple swallows. On each swallow operation any bacteria $i$ can swallow any bacteria $j$ if $a_i>a_j$ and $a_i\le a_j+K$. The swallow operations go one after another.

For example, the sequence of bacteria sizes $a=[101,53,42,102,101,55,54]$ and $K=1$. The one of possible sequences of swallows is: $[101,53,42,102,\underset{–}{101},55,54]$ $\to$ $[101,\underset{–}{53},42,102,55,54]$ $\to$ $[\underset{–}{101},42,102,55,54]$ $\to$ $[42,102,55,\underset{–}{54}]$ $\to$ $[42,102,55]$. In total there are $3$ bacteria remained in the Petri dish.

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 $n$ and $K$ ($1\le n\le 2\cdot {10}^5$, $1\le K\le {10}^6$) — number of bacteria and intergalactic constant $K$.

The second line contains $n$ space separated integers $a_1,a_2,\dots ,a_n$ ($1\le a_i\le {10}^6$) — sizes of bacteria you have.

Output

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

Examples

input

Copy

7 1
101 53 42 102 101 55 54

output

Copy

3

input

Copy

6 5
20 15 10 15 20 25

output

Copy

1

input

Copy

7 1000000
1 1 1 1 1 1 1

output

Copy

7

Note

The first example is clarified in the problem statement.

In the second example an optimal possible sequence of swallows is: $[20,15,10,15,\underset{–}{20},25]$ $\to$ $[20,15,10,\underset{–}{15},25]$ $\to$ $[20,15,\underset{–}{10},25]$ $\to$ $[20,\underset{–}{15},25]$ $\to$ $[\underset{–}{20},25]$ $\to$ $\left[25\right]$.

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

题意：你有一个培养皿，里面n个细菌，每个细菌有自己的大小，大的细菌能吃掉小的细菌，自身大小不变，前提是大细菌只能吃掉小细菌加k系数大于大细菌才行。这句话就是条件  only if $a_i>a_j$ and $a_i\le a_j+\mathrm{K，求出培养皿中最少还有多少细菌。}$

题解：贪心  排个序后，然后依次比较小细菌是否加k大于大的细菌，满足的话，就总个数--。

#include<bits/stdc++.h>
using namespace std;
int n,k,a[200100],j;
int main()
{
    cin>>n>>k;
    for(int i=0; i<n; i++)
        cin>>a[i];
    sort(a,a+n);
    int ans=n;
    for(int i=0; i<n; i++)
        while(a[j]<a[i])
        {
            if(a[i]<=a[j]+k) ans--;
            j++;
        }
    cout<<ans<<endl;
    return 0;
}

n,k=map(int,input().split())
a=sorted(map(int,input().split()))
j=0
for i in a:
    while a[j]<i:
        if i<=a[j]+k:
            n-=1
        j+=1
print(n)