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
2016-07-12 19:18:18
Got WA because of INT.
2016-07-12 09:18:03
good application of BIT
2016-07-03 12:39:13
why is the problem tagged as shortest path?
2016-06-28 06:05:41
You would need to fully understand Merge sort/BIT to even start with this ... if you choose it to do via merge sort , this video will finally tell you the logic you need to use to count the number of inversions https://www.youtube.com/watch?v=hqeoAIryJOc
2016-06-26 20:07:32
Use long for number of inversions
Costed me so many WAs -.-
2016-06-22 15:18:30
Don't use STL set (gives TLE) use MergeSort
2016-06-21 22:43:19
CLRS rocks :-) AC in one go !
2016-06-14 20:01:23
inversions are hidden in mergesort
2016-06-11 09:57:54
just make everything long long and leave a line after each test case by ur own
2016-06-01 11:21:19
use long long for no.of inversions counter (else it would result WA)
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