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
 
如果差距在k及以内,两者都有赢的可能。如果差距大于k,大者赢。
因此找出两两差距都在k以内的大数有几个即可。
 
 #include <iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<cmath>
#include<string>
#include<map>
#include<vector>
using namespace std; int a[];
bool cmp(int x,int y)
{
return x>y;
} int main()
{
int T,n,k;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&k);
for(int i=;i<n;i++)
scanf("%d",&a[i]);
sort(a,a+n,cmp);
int i;
for(i=;i<n-;i++)
{
if(a[i]-a[i+]>k)
break;
}
printf("%d\n",i+);
}
return ;
}
 
 

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