D. Balanced Round

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are the author of a Codeforces round and have prepared n� problems you are going to set, problem i� having difficulty ai��. You will do the following process:

• remove some (possibly zero) problems from the list;
• rearrange the remaining problems in any order you wish.

A round is considered balanced if and only if the absolute difference between the difficulty of any two consecutive problems is at most k� (less or equal than k�).

What is the minimum number of problems you have to remove so that an arrangement of problems is balanced?

Input

The first line contains a single integer t� (1≤t≤10001≤�≤1000) — the number of test cases.

The first line of each test case contains two positive integers n� (1≤n≤2⋅1051≤�≤2⋅105) and k� (1≤k≤1091≤�≤109) — the number of problems, and the maximum allowed absolute difference between consecutive problems.

The second line of each test case contains n� space-separated integers ai�� (1≤ai≤1091≤��≤109) — the difficulty of each problem.

Note that the sum of n� over all test cases doesn't exceed 2⋅1052⋅105.

Output

For each test case, output a single integer — the minimum number of problems you have to remove so that an arrangement of problems is balanced.

Example

input

Copy

7

5 1

1 2 4 5 6

1 2

10

8 3

17 3 1 20 12 5 17 12

4 2

2 4 6 8

5 3

2 3 19 10 8

3 4

1 10 5

8 1

8 3 1 4 5 10 7 3

output

Copy

2
0
5
0
3
1
4

Note

For the first test case, we can remove the first 22 problems and construct a set using problems with the difficulties [4,5,6][4,5,6], with difficulties between adjacent problems equal to |5−4|=1≤1|5−4|=1≤1 and |6−5|=1≤1|6−5|=1≤1.

For the second test case, we can take the single problem and compose a round using the problem with difficulty 1010.

解题说明：此题是一道数学题，分析后能发现直接对数列排序，然后再遍历一遍找出符合两个数之间差 <= k 的元素数量是多少， 再用总数量减去符合两个数之间差 <= k 的元素最长连续数量。

#include
using namespace std;
int t, n, k, ans, temp, a[200000];
int main() 
{
	cin >> t;
	while (t--)
	{
		cin >> n >> k;
		for (int i = 0; i < n; i++)
		{
			cin >> a[i];
		}
		sort(a, a + n);
		temp = 1;
		ans = 1;
		for (int i = 1; i < n; i++) 
		{
			if (a[i] - a[i - 1] <= k)
			{
				temp++;
			}
			else
			{
				temp = 1;
			}
			ans = max(ans, temp);
		}
		cout << n - ans << '\n';
	}
	return 0;
}