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E - Chinese Restaurant (Three-Star Version)

Score : $500$ points

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.

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.

4
1 2 0 3

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 $2$.

3
0 1 2

0

10
3 9 6 1 7 2 8 0 5 4

20