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
2018-03-16 12:15:54
hint : : use BIT or Merge
seriously invested more than 4 hours to come up with sol.. nd debugged a lot of times to understand the flow of code nd finally ac in 1go !!! gys try something new rather than merge ..go for hacker-earth tut for this.

Last edit: 2018-03-16 19:55:46
2018-03-16 12:01:15
I'm getting WA!!! Somebody please help.

<snip>

Last edit: 2022-06-26 16:00:27
2018-03-16 07:15:19
@addy1397 There are n distinct numbers in the problem, none of them are repeated. Some of your random inputs are invalid.
2018-02-26 22:01:11
Use BigInteger (number of inversions) for java else you will get WA.

Last edit: 2018-02-26 22:02:24
2018-01-14 14:14:47
my 100th :) using BIT
2017-12-31 17:32:15
Make sure to use long long in c++
2017-12-28 11:14:17
I used segment tree.
But gives TLE...
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!
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