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E. Little Elephant and Inversions

time limit per test

2 secondsmemory limit per test

256 megabytesinput

standard inputoutput

standard outputThe Little Elephant has array *a*, consisting of *n* positive integers, indexed from 1 to *n*. Let's denote the number with index *i* as *a*_{i}.

The Little Elephant wants to count, how many pairs of integers *l* and *r* are there, such that 1 ≤ *l* < *r* ≤ *n* and sequence *b* = *a*_{1}*a*_{2}... *a*_{l}*a*_{r}*a*_{r + 1}... *a*_{n} has no more than *k* inversions.

An inversion in sequence *b* is a pair of elements of the sequence *b*, that change their relative order after a stable sorting of the sequence. In other words, an inversion is a pair of integers *i* and *j*, such that 1 ≤ *i* < *j* ≤ |*b*| and *b*_{i} > *b*_{j}, where |*b*| is the length of sequence *b*, and *b*_{j} is its *j*-th element.

Help the Little Elephant and count the number of the described pairs.

Input

The first line contains two integers *n* and *k* (2 ≤ *n* ≤ 10^{5}, 0 ≤ *k* ≤ 10^{18}) — the size of array *a* and the maximum allowed number of inversions respectively. The next line contains *n* positive integers, separated by single spaces, *a*_{1}, *a*_{2}, ..., *a*_{n} (1 ≤ *a*_{i} ≤ 10^{9}) — elements of array *a*.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.

Output

In a single line print a single number — the answer to the problem.

Examples

Input

3 1

1 3 2

Output

3

Input

5 2

1 3 2 1 7

Output

6

Codeforces (c) Copyright 2010-2018 Mike Mirzayanov

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