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
2017-06-19 21:54:15
I am getting TLE in my merge_sort approach. Plz help me how can I make my code efficient.
Here is my Code link: <snip>
Thanks in advance :)

Last edit: 2022-06-26 16:04:40
2017-06-19 07:24:22
Easy one AC in one go
2017-06-12 18:05:30
AC in one go!!
Only Merge Sort..
2017-06-02 09:48:56
AC in one go))))
2017-05-25 04:38:39
Core concept of sorting algorithm..
2017-05-02 14:53:24
thanks @fane_faiz :)
2017-04-10 19:42:03
Merge Sort->"ACIN1GO"
2017-04-05 20:10:49
Segment Tree+Merge Sort AC in 1 go :)
2017-04-05 14:12:46
If BST implementation gets AC, then the test cases are weak because the running time for BST is O(nh), which is O(n^2) in worst case. Basically it should fail if the sequence is sorted.
2017-03-23 18:42:59
solved using binary search tree..AC in one go
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