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

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
2021-06-30 06:33:50
Getting TLE, even after using merge sort algorithm and scanf/printf. Any idea. Why it is so
2021-06-23 16:15:35
Can be easily solved using the Merge Sort function. Just add some checks while merging and you are good to go.
2021-06-04 10:02:16
Similar problems using Ordered_set:
Points on Line(CF)
Sliding Median(CSES)
Sliding Cost(CSES)
Inversion Count(SPOJ)
2021-05-15 10:21:11
A good problem to learn PBDS if you use C++
https://www.geeksforgeeks.org/ordered-set-gnu-c-pbds/
2021-04-09 10:29:22
Ac in one go using segment tree
2021-03-31 07:07:42
modified merge sort on passing by reference and using printf/scanf (instead of cin/cout ) gets accepted.
Apparently differences are too big between cin/cout and scanf/printf in SPOJ.
2021-03-29 14:22:52
Solved using Fenwick Tree !
2021-03-16 08:18:55
Finally get AC by using binary search tree
2021-02-24 12:25:00
Return type long long vs int tripped me up.

Last edit: 2021-02-24 12:25:12
2021-02-03 19:32:52
A really good sum. But wasted so much time in tle. Pass by reference to remove tle.
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