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
MAYANK NARULA: 2015-07-12 23:16:54

Aah well somethings never change! Just don't forget to use long long..
And while calculating middle index dont overflow.
Plus! i made a silly mistake to take MAX as 10^6 rather than 10^7

Last edit: 2015-07-12 23:17:43
rini22: 2015-07-11 16:29:13

use long long

chin: 2015-07-02 10:53:20

AC in 2nd attempt!!....A very good application of Mergesort!!

mohit: 2015-06-28 18:32:31

mergesort .. AC after 4 WA!!! give special attention to type...

THECOLDONE: 2015-06-24 14:14:31

after reading BIT done it by merge algo
ac in 1st go

Aditya Kumar: 2015-06-24 09:18:21

Be careful with overflows

Sahil: 2015-06-20 15:42:51

You only need the inversion count in long long (guess why? look at the worst case for this problem).
rest everything(size of array, array elements, all index) can be int.

[Mayank Pratap]: 2015-06-20 05:22:08

After months finally AC :)

kartikay singh: 2015-06-18 18:56:34

after lots of TLEs
finally AC:)

r0bo_dart: 2015-06-13 13:35:15

HINT: Dont use vector for this problem with your custom designed insert. That will slow things out. Use array instead. Cost me 4 TLEs... phew. Else Merge Sort works nice. and yeah the long long thingy too.


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