Little Petya likes points a lot. Recently his mom has presented him n points lying on the line OX. Now Petya is wondering in how many ways he can choose three distinct points so that the distance between the two farthest of them doesn't exceed d.

Note that the order of the points inside the group of three chosen points doesn't matter.

The first line contains two integers: n and d (1 ≤ n ≤ 105; 1 ≤ d ≤ 109). The next line contains n integers x1, x2, ..., xn, their absolute value doesn't exceed 109 — the x-coordinates of the points that Petya has got.

It is guaranteed that the coordinates of the points in the input strictly increase

Print a single integer — the number of groups of three points, where the distance between two farthest points doesn't exceed d.

Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64dspecifier.

4 3
1 2 3 4

4

4 2
-3 -2 -1 0

2

5 19
1 10 20 30 50

1

In the first sample any group of three points meets our conditions.

In the seconds sample only 2 groups of three points meet our conditions: {-3, -2, -1} and {-2, -1, 0}.

In the third sample only one group does: {1, 10, 20}.

題意：

給出n和k，然後給出一個大小為 n 的集合，從集合中取出3個數，
最大數與最小數的差值小於k的方式有多少種。

程式碼如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
#define LL __int64
int main()
{
    int n;
    LL d;
    LL a[100047];
    while(~scanf("%d%I64d",&n,&d))
    {
        for(int i = 0; i < n; i++)
        {
            scanf("%I64d",&a[i]);
        }
        LL ans = 0;
        for(int i = 1; i < n; i++)
        {
            LL tt = i-(lower_bound(a,a+i,a[i]-d)-a)-1;
            ans+=tt*(tt+1)/2;//C n 2
        }
        printf("%I64d\n",ans);
    }
    return 0;
}
/*
4 3
1 2 3 4
4 2
-3 -2 -1 0
5 19
1 10 20 30 50
*/