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 $*d*$ ($1 ≤ *n* ≤ 10^{5}; 1 ≤ *d* ≤ 10^{9}$). The next line contains $*n*$ integers $*x*_{1}, *x*_{2}, ..., *x*_{*n*}$, their absolute value doesn't exceed $10^{9}$ — 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 %I64d specifier.

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}.