We've got no test cases. A big olympiad is coming up. But the problemsetters' number one priority should be adding another problem to the round.

The diameter of a multiset of points on the line is the largest distance between two points from this set. For example, the diameter of the multiset {1, 3, 2, 1}

Diameter of multiset consisting of one point is 0.

You are given n points on the line. What is the minimum number of points you have to remove, so that the diameter of the multiset of the remaining points will not exceed d?

The first line contains two integers n and d (1 ≤ n ≤ 100, 0 ≤ d ≤ 100) — the amount of points and the maximum allowed diameter respectively.

The second line contains n space separated integers (1 ≤ xi ≤ 100) — the coordinates of the points.

Output a single integer — the minimum number of points you have to remove.

3 1
2 1 4

1

3 0
7 7 7

0

6 3
1 3 4 6 9 10

3

題意：給你n個數，和一個k，這n個數的最大差值不能超過k，問你最少需要刪除幾個數滿足要求。

思路：逆向思維，我們找最長符合要求的長度，n減去這個長度就是需要刪除的個數。由於一定是連續的，所以n^2列舉。

#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll a[110],n,d,mmax,ans;
int main()
{
    while(~scanf("%lld%lld",&n,&d))
    {
        for(ll i=1;i<=n;i++)scanf("%lld",&a[i]);
        sort(a+1,a+n+1);
        ans=-1;
        for(ll i=1;i<=n;i++)
        {
            mmax=1;
            for(ll j=i+1;j<=n;j++)
            {
                if(a[j]-a[i]<=d)mmax++;
                else break;
            }
            if(mmax>ans)ans=mmax;
        }
        printf("%lld\n",n-ans);
    }
    return 0;
}