Problem9706--ABC190 —— F - Shift and Inversions

9706: ABC190 —— F - Shift and Inversions

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Time Limit : 1.000 sec  Memory Limit : 512 MiB

Description

Given is a sequence $A = [a_0, a_1, a_2, \dots, a_{N-1}]$ that is a permutation of $0, 1, 2, \dots, N - 1$.  
For each $k = 0, 1, 2, \dots, N - 1$, find the inversion number of the sequence $B = [b_0, b_1, b_2, \dots, b_{N-1}]$ defined as $b_i = a_{i+k\ \bmod\ N}$.

What is inversion number? The inversion number of a sequence $A = [a_0, a_1, a_2, \dots, a_{N-1}]$ is the number of pairs of indices $(i, j)$ such that $i < j$ and $a_i > a_j$.

Input

Input is given from Standard Input in the following format:

```
$N$
$a_0$ $a_1$ $a_2$ $\cdots$ $a_{N-1}$
```

Output

Print $N$ lines.  
The $(i + 1)$-th line should contain the answer for the case $k = i$.

Constraints

-   All values in input are integers.
-   $2 ≤ N ≤ 3 \times 10^5$
-   $a_0, a_1, a_2, \dots, a_{N-1}$ is a permutation of $0, 1, 2, \dots, N - 1$.

Sample 1 Input

4
0 1 2 3

Sample 1 Output

0
3
4
3
We have A=[0,1,2,3].
For k=0, the inversion number of B=[0,1,2,3] is 0.
For k=1, the inversion number of B=[1,2,3,0] is 3.
For k=2, the inversion number of B=[2,3,0,1] is 4.
For k=3, the inversion number of B=[3,0,1,2] is 3.

Sample 2 Input

10
0 3 1 5 4 2 9 6 8 7

Sample 2 Output

9
18
21
28
27
28
33
24
21
14

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