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.)

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.

Input

Input is given from Standard Input in the following format:

$N$ $p_0$ $\\ldots$ $p_{N-1}$

Output

Print the answer.

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