Codeforces Round #466 (Div. 2) A. Points on the line

A. Points on the line

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} is 2.

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?

Input

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

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

Examples

input

Copy

3 1
2 1 4

output

1

input

Copy

3 0
7 7 7

output

0

input

Copy

6 3
1 3 4 6 9 10

output

3

题意：给你n个数，和一个k，这n个数的最大差值不能超过k，问你最少需要删除几个数满足要求。

思路：逆向思维，我们找最长符合要求的长度，n减去这个长度就是需要删除的个数。由于一定是连续的，所以n^2枚举。

#include
#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;
}