CF940A.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?

我们没有测试用例。一场大型奥林匹克竞赛即将举行。但出题人的首要任务是为该轮比赛再添加一道题目。

一个数轴上点的多重集的直径，是指该集合中任意两点之间的最大距离。例如，多重集 $\{1,\,3,\,2,\,1\}$ 的直径为 $2$。

仅含一个点的多重集的直径为 $0$。

给定数轴上的 $n$ 个点。你需要至少删除多少个点，才能使得剩余点构成的多重集的直径不超过 $d$？

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 ≤ x__i ≤ 100) — the coordinates of the points.

第一行包含两个整数 $n$ 和 $d$（$1 \leq n \leq 100$，$0 \leq d \leq 100$）—— 分别表示点的数量和允许的最大直径。

第二行包含 $n$ 个用空格分隔的整数（$1 \leq x_i \leq 100$）—— 表示各点的坐标。

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

输出一个整数——你需要移除的最少点数。

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.

在第一个测试用例中，最优策略是移除坐标为 $4$ 的点。剩余点的坐标为 $1$ 和 $2$，因此直径等于 $2 - 1 = 1$。

在第二个测试用例中，直径等于 $0$，因此无需移除任何点。

在第三个测试用例中，最优策略是移除坐标为 $1$、$9$ 和 $10$ 的点。剩余点的坐标为 $3$、$4$ 和 $6$，因此直径等于 $6 - 3 = 3$。