Problem Statement

Person $0, Person 1, \ldots$, and Person (N-1) are sitting around a turntable in counterclockwise order, evenly spaced. Dish p_i is in front of Person i on the table.
You may perform the following operation 0 or more times:

• Rotate the turntable by one N-th of a counterclockwise turn. The dish that was in front of Person i right before the rotation is now in front of Person $(i+1) \bmod N$.

When you are finished, Person i gains frustration of k, where k is the minimum integer such that Dish i is in front of either Person $(i-k) \bmod N$ or Person $(i+k) \bmod N$.
Find the minimum possible sum of frustration of the N people.
What is $a \bmod m$? For an integer a and $a positive integer m, a \bmod m$ denotes the integer x between 0 and (m-1) (inclusive) such that (a-x) is a multiple of m. (It can be proved that such x is unique.)

Input

Input is given from Standard Input in the following format:
N
$p_0 \ldots p_{N-1}$

Output

Print the answer.

Constraints

• $3 \leq N \leq 2 \times 10^5$
• $0 \leq p_i \leq N-1$
• $p_i \neq p_j$ if $i \neq j$.
• All values in input are integers.

Sample Input 1

4
1 2 0 3

Sample Output 1

2

The figure below shows the table after one operation.

Here, the sum of their frustration is 2 because:

• Person 0 gains a frustration of 1 since Dish 0 is in front of Person 3\ $(=(0-1) \bmod 4)$;
• Person 1 gains a frustration of 0 since Dish 1 is in front of Person 1\ $(=(1+0) \bmod 4)$;
• Person 2 gains a frustration of 0 since Dish 2 is in front of Person 2\ $(=(2+0) \bmod 4)$;
• Person 3 gains a frustration of 1 since Dish 3 is in front of Person 0\ $(=(3+1) \bmod 4)$.

We cannot make the sum of their frustration less than 2, so the answer is 2.

Sample Input 2

3
0 1 2

Sample Output 2

0

Sample Input 3

10
3 9 6 1 7 2 8 0 5 4

Sample Output 3

20

见题面 Input 部分

见题面 Output 部分