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] ≤ 107). 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
[Lakshman]: 2014-11-12 12:19:45

@Abi you have declared count as long long and your mergesort is returning int , So declare mergesort and merge long long instead. Hope this helps.

Last edit: 2014-11-12 12:25:19
Abi: 2014-11-12 06:59:38

I dont know ....why iam getting WA :(
pls anyone help
<snip>

Last edit: 2022-06-26 16:05:00
surayans tiwari(http://bit.ly/1EPzcpv): 2014-11-08 12:29:10

acce using merge ,, got acc using fast i/o with bit

Last edit: 2014-11-08 12:44:38
Vikas: 2014-10-12 15:39:58

I am getting WA even after checking my code many times.Please help.
My id is 12607511

vikax: 2014-09-25 13:49:41

3 RTE coz of a silly mistake...... finally ac

Siya: 2014-09-23 12:49:05

AC in first chance :)

saumya: 2014-09-14 14:09:45

same as http://www.spoj.com/problems/CODESPTB/

just use unsigned long long int.. as said.. it cost me WA !! :/

Gaurav Ahirwar: 2014-08-30 13:29:12

0.70 secs AC! :D

ankur manchanda: 2014-07-22 10:18:51

declare variable count as long long int.
It cost me two WA.
As in worst case count =2*10^5(2*10^5+1)/2;

Last edit: 2014-07-22 10:21:54
Anmol Pandey: 2014-07-18 17:30:53

Take array size 10^7+1 to remove SIGSEGV error


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