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Codeforces Round 153 (Div. 1) |
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Finished |

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binary search

combinatorics

two pointers

*1300

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A. Points on Line

time limit per test

2 secondsmemory limit per test

256 megabytesinput

standard inputoutput

standard outputLittle 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.

Examples

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}.

Codeforces (c) Copyright 2010-2023 Mike Mirzayanov

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