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

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

hide comments
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.
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.
2017-08-25 16:36:54
Merge sort implementation in a innovative way.............
2017-08-21 13:00:29
AC in one go using trie
2017-07-30 17:01:52
Comments are like GOT spoilers. Never look at them unless you are done ;)
2017-07-24 13:50:35
use merge sort and AC!!!
2017-07-01 13:37:46
AC in 1
Just MergeSort
2017-06-30 18:26:07
Binary tree, easy :P
2017-06-26 09:40:03
took me 2 hours to figure out long long mistake !!!
2017-06-19 21:54:15
I am getting TLE in my merge_sort approach. Plz help me how can I make my code efficient.
Here is my Code link: <snip>
Thanks in advance :)

Last edit: 2022-06-26 16:04:40
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