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
2017-12-18 23:39:12
Although I solved it using merge sort but can anyone help me in its bitmask approach?

Last edit: 2017-12-21 13:01:48
2017-12-18 21:41:05
easy BIT...int gives WA..use long long
2017-12-18 10:00:24
came to know a creative of using fenwick tree!
2017-12-01 05:07:40
Try Fenwick Tree

Random testcases
Input
5
6
1 2 3 4 5 6
8
5 1 4 2 6 2 6 2
5
4 1 2 4 2
7
1 4 2 5 2 56 2
1
1

output

0
11
4
6
0


2017-11-13 12:41:27
BST approach using STL gives TLE! but mergesort works :)

Last edit: 2017-11-13 14:54:21
2017-10-23 21:40:05
tags are clickbait
2017-10-16 11:35:56 harkirat
1

Last edit: 2017-10-16 11:36:08
2017-10-05 00:27:52
Use Long long int if you don't want WA!
2017-10-04 20:30:53
Mergesort does the job
2017-09-30 20:20:16
If you want to do it in orthrodox manner do it by merge sort.Its just modified mergesort.It can be solved using BIT concept too.There is a bit mask approach too but it gives TLE.AC in 4th go....
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