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

A. Points on the line

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

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, 4and 6, so the diameter will be equal to 6 - 3 = 3.

题意大概就是，我给出n个数和他们之间最大的差d，问你最少需要删掉几个数，可以让他们之间最大的差小于等于d。

这个题目我一开始的思路是用vector暴力，sort之后因为vector可以方便的去头去尾，每次去一次头/尾就判断一次，结果发现不太对劲....万一有一个1 2 3 98 99 100 d=5，那不就凉了......

后来发现其实这个题反正就100个数...那我二重循环直接暴力也可以，sort之后找到符合相邻两项小于等于d的最大序列就好了，然后维护一个sum。

#include<bits/stdc++.h>

using namespace std;
int n,d;
int a[105];

int main()
{
    cin>>n>>d;
    for(int i=0; i<n; i++)
        cin>>a[i];
    sort(a,a+n);
    int cnt;
    int sum=-1;
    for(int i=0; i<n; i++)
    {
        cnt=1;
        for(int j=i+1; j<n; j++)
        {
            if(a[j]-a[i]<=d) cnt++;
        }
        sum=max(sum,cnt);
    }
    cout<<n-sum<<endl;
    return 0;
}