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
dushyant7917: 2017-11-13 12:41:27

BST approach using STL gives TLE! but mergesort works :)

Last edit: 2017-11-13 14:54:21
aimbot: 2017-10-23 21:40:05

tags are clickbait

harkirat: 2017-10-16 11:35:56

1

Last edit: 2017-10-16 11:36:08
ashutosh450: 2017-10-05 00:27:52

Use Long long int if you don't want WA!

hitesh87: 2017-10-04 20:30:53

Mergesort does the job

vishesh197: 2017-09-30 20:20:16

If you want to do it in orthrodox manner do it by merge sort.Its just modified mergesort.It can be solved using BIT concept too.There is a bit mask approach too but it gives TLE.AC in 4th go....

Dinu Bhai: 2017-09-19 20:06:02

dont forget to print the answer in a new line...

Last edit: 2017-09-19 20:06:19
anmoldua: 2017-09-14 20:44:47

fenwick tree easy

taksh: 2017-09-08 14:19:01

How to you solve this? Someone please send solution code for this. Been trying for the last 3 days.

polkerty: 2017-09-05 04:13:19

So my solution is really weird (respect to everyone who saw the merge sort). Basically I used 3 vectors and some carefully chosen constants to get the answer by sorting progressively smaller pieces of the array, and then using a bunch of binary searches to find the values I wanted.


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