940A. Points on the line(思维逆向)

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本文探讨了一道关于在坐标轴上删除部分点以确保剩余点集直径不超过特定值的问题。介绍了解题思路，并提供了一个使用C++实现的具体算法示例。

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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 ≤ x_i ≤ 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

Input

Copy

3 0
7 7 7

Output

Input

Copy

6 3
1 3 4 6 9 10

Output

Note

In the first test case the optimal strategy is to remove the point with coordinate 4. The remaining points will have coordinates 1 and 2, so the diameter will be equal to 2 - 1 = 1.

In the second test case the diameter is equal to 0, so its is unnecessary to remove any points.

In the third test case the optimal strategy is to remove points with coordinates 1, 9 and 10. The remaining points will have coordinates 3, 4 and 6, so the diameter will be equal to 6 - 3 = 3.

题意：问最少删除几个数字，使得剩余数字的最大值 - 最小值 < = m；

思路：思维逆向，求符合题意的最多数字

 #include<bits/stdc++.h>
using namespace std;
const int N = 105;
int arr[N];
int main(){
	int n,m;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
	   scanf("%d",&arr[i]);
	sort(arr+1,arr+n+1);
	int ans=0;
	for(int i=1;i<=n;i++){
		for(int j=i;j<=n;j++)
		if(arr[j]-arr[i]<=m)
		   ans=max(ans,j-i+1);
	}
	printf("%d\n",n-ans);
	return 0;
}