Points on Line

 Points on Line

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.

Input

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.

Output

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.

Sample test(s)

input

4 3
1 2 3 4

output

4

input

4 2
-3 -2 -1 0

output

2

input

5 19
1 10 20 30 50

output

1

Note

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}.

--------------------------------------------------------editorial----------------------------------------------------------------

Let's select the rightmost point of our triplet. In order to do this we can iterate over all points in ascending order of their X-coordinate. At the same time we'll maintain a pointer to the leftmost point which lays on the distance not greater than d from the current rightmost point. We can easily find out the number of points in the segment between two pointers, excluding the rightmost point. Let's call this number k. Then there exist exactly k * (k - 1) / 2 triplets of points with the fixed rightmost point. The only thing left is to sum up these values for all rightmost points.

----------------------------------------------------code------------------------------------------------------------------------

#include<bits/stdc++.h>

using namespace std;

typedef long long int lli;

typedef long int li;

 vector<lli> v;

  lli ans=0;

  

int main()

{

int n,d;

 cin>>n>>d;

 for(int i=0;i<n;i++)

  {

  lli a;

   cin>>a;

   v.push_back(a);

  }

  sort(v.begin(),v.end());

  for(int i=0;i<n-2;i++)

   {

    // cout<<" i "<<i<<endl;

        lli maxi;

    lli pre=v[i];

    // if(pre <0)

      maxi=pre+d;

       vector<lli> :: iterator it;

       it=upper_bound(v.begin()+i+2,v.end(),maxi);

       if(it==v.end() || *it-d>v[i]) it--;

      lli sets=it-v.begin()-i;

       // cout<<sets<<endl;

       ans+=(sets*(sets-1))/2;

      //  cout<<ans<<endl;

     

}

  

cout<<ans<<endl;

}