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
dennislo: 2018-06-09 17:00:02

Use long in Java to avoid WA

futile55: 2018-06-09 11:35:55

Merge sort and AC :)

a000: 2018-05-24 02:27:40

The test cases are very weak... I know my code is wrong still I went for a try and it got accepted.

devarshi09: 2018-05-21 14:09:12

Its easy to think of a solution if one knows merge sort tree. Got to think more for BIT!

aditya12legend: 2018-03-30 19:43:28

Last edit: 2018-03-30 19:58:20
a2j007: 2018-03-20 17:55:00

Segment tree ->0.28 :)

shauryauppal: 2018-03-18 08:37:26

BIT with long long is the key to success.

jmr99: 2018-03-16 19:57:39

how to solve this using #graph-theory #number-theory #shortest-path pls somebody share the idea .

mranderson: 2018-03-16 19:52:03

use long long int for the count

jmr99: 2018-03-16 16:24:21

Last edit: 2018-03-16 19:52:08

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