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

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
2016-02-27 13:10:49 arthur
Can be solved using segment, BIT, range trees. Also try and solve LIS using segment trees. All solutions should be O(nlgn)
2016-02-16 06:47:14 Devashish Mathur
Cheated.. but Loved it..
2016-01-28 13:01:16
If this is the first question you are trying to do using BIT, then please try some other questions of BIT before this one.... It took me quite some while to figure out the logic behind this question's BIT implementation.... :)
2016-01-24 20:49:14
Thanks @Kartik. I would have toiled for hours to figure out the WA.
2016-01-21 19:52:28
nice problem !!!
2016-01-11 00:41:57 Papai


Last edit: 2016-01-11 00:52:52
2016-01-05 19:46:17
use long long int(better...for all variables)....and better...try using merge sort algorithm!
2016-01-05 14:59:16
2nd test-case n= 5-> 2 3 8 6 1
inversion pairs are ( 2 , 1 ), (3,1) , (8 , 6) , (8 , 1) , (6 , 1) .
2015-12-29 18:04:11
For all those struggling to understand how exactly merge sort will be useful should refer this
http://www.geeksforgeeks.org/counting-inversions/
2015-12-26 06:55:33 Kartik
keep n long long, cost me a WA
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