Problem Description

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:

A wrestling match will be held tomorrow. n players will take part in it. The ith player’s strength point is ai.

If there is a match between the ith player plays and the jth player, the result will be related to |ai−aj|. If |ai−aj|>K, the player with the higher strength point will win. Otherwise each player will have a chance to win.

The competition rules is a little strange. Each time, the referee will choose two players from all remaining players randomly and hold a match between them. The loser will be be eliminated. After n−1 matches, the last player will be the winner.

Now, Yuta shows the numbers n,K and the array a and he wants to know how many players have a chance to win the competition.

It is too difficult for Rikka. Can you help her?

Input

The first line contains a number t(1<=t<=100), the number of the testcases. And there are no more than 2 testcases with n>1000.

For each testcase, the first line contains two numbers n,K(1<=n<=105,0<=K<109).

The second line contains n numbers ai(1<=ai<=109).

Output

For each testcase, print a single line with a single number – the answer.

Sample Input

2 
5 3 
1 5 9 6 3 
5 2 
1 5 9 6 3

Sample Output

5 
1

题目大意：

摔跤比赛将于明天举行。运动员将参加。随着玩家的实力是AI。 
如果有与玩家之间，金天海球员比赛，结果将是| AI−AJ |相关。如果| AI−AJ |＞K，与高强度点的玩家将赢得。否则，每个玩家将有机会赢得。 
比赛的规则有点奇怪。每一次，裁判都会随机从剩下的球员中选出两名球员，并在他们之间进行比赛。失败者将被淘汰。N−1场比赛后，最后的球员将是赢家。 
现在，Yuta显示数N、K和阵列，他想知道有多少玩家有机会赢得比赛。 
把 a_ia 
​i 
​​ 从大到小排序，那么第 ii 强人要获胜，最优情况下是最强的人输给了第二强的人，第二强的人输给了第三强的人，以此类推。因此只需要判断排序后 max_{j

c++

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int N;
    cin>>N;
    while(N--)
    {
        int n,k;
        int a[500001];
        cin>>n>>k;
        for (int i=0; i<n; i++)
        {
            cin>>a[i];
        }
        sort(a,a+n);
        int ans=0;
        int flag=0;
        for (int i=n-1; i>0; i--)
        {
            if (abs(a[i]-a[i-1])<=k)
            {
                ans += 1;
                //cout<<"!!!!"<<endl;
            }
            else
            {
                ans += 1;
                flag=1;
                break;
            }
        }
            if (abs(a[0]-a[1])<=k&&flag==0)
                ans++;
        cout<<ans<<endl;
    }
}