INVCNT - Inversion Count


Let A[0...n - 1] be an array of n distinct positive integers. If i < j and A[i] > A[j] then the pair (i, j) is called an inversion of A. Given n and an array A your task is to find the number of inversions of A.

Input

The first line contains t, the number of testcases followed by a blank space. Each of the t tests start with a number n (n <= 200000). Then n + 1 lines follow. In the ith line a number A[i - 1] is given (A[i - 1] <= 10^7). The (n + 1)th line is a blank space.

Output

For every test output one line giving the number of inversions of A.

Example

Input:
2

3
3
1
2

5
2
3
8
6
1


Output:
2
5

hide comments
Himanshu Garg: 2015-08-11 13:26:21

Learn MergeSort & GO AHEAD...YOYO :)

Anh Quan: 2015-07-30 18:04:49

use long long u will get AC :) if use int u will get WA out!

SUBHAJIT GORAI: 2015-07-29 12:52:05

use long long to store the result ...

just_code21: 2015-07-25 05:15:36

stop putting useless comments to confuse....!!

anando_du: 2015-07-23 21:33:09

used BIT + Binary search (for compression) got AC .
long long caused one WA :v

aman mahansaria: 2015-07-21 09:26:53

finally AC :) pls use long long instead of int

Komal: 2015-07-19 11:51:23

for the number of inversions do use a 64 bit integer(long in java)

Last edit: 2015-07-19 11:51:55
ice_cold: 2015-07-14 17:55:54

i got it correct for one test case but i dont know how to apply it for t test cases , pls give a hint .
p.s. i am using c++ and just started coding and i am 16.

MAYANK NARULA: 2015-07-12 23:16:54

Aah well somethings never change! Just don't forget to use long long..
And while calculating middle index dont overflow.
Plus! i made a silly mistake to take MAX as 10^6 rather than 10^7

Last edit: 2015-07-12 23:17:43
rini22: 2015-07-11 16:29:13

use long long


Added by:Paranoid Android
Date:2010-03-06
Time limit:3.599s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: PERL6