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
2023-03-11 09:54:46
This problem is solvable using:
1- Divide & Conquer (Merge Sort)
2- Segment tree
3- Fenweick tree (BIT)
4- Self-Balancing BST (indexed_set in C++)
5- Tries (Idk how does this work)
2022-11-07 14:44:15
I solved it using a binary tree :)
2022-08-08 17:59:27
Can this problem can be solved using Segment tree ?
2022-01-11 20:03:21
why is my solution accepted in CPP 14 but giving WA in C++ ?
2022-01-03 15:17:13
you have given that arr[i-1]<=10^7 , but it causes int overflow, when i used long long it got submitted, please correct it.

Last edit: 2022-01-03 15:17:27
2021-12-27 07:23:11
AC in one GO Using ordered set
2021-11-13 08:37:16
<snip>

[NG]: Quit posting code / solutions / spoilers or you'll see your commenting privilege removed.

Last edit: 2021-11-13 09:42:38
2021-11-03 13:52:57 4444
@sabkx thanks a lot for the long long int comment
2021-10-03 07:06:22
EZ AC in second go! use long long instead of int. Merge sort works with 0.11 second
2021-08-19 06:33:52
Fenwick Tree :D, return type long long int
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