P16842 [GKS 2021 #B] Longest Progression
Description
In [Kick Start $2020$ Round $E$](https://www.luogu.com.cn/problem/P16763) (you do not need to know anything about the previous problem to solve this one) Sarasvati learned about arithmetic arrays. An arithmetic array is an array that contains at least $2$ integers and the differences between consecutive integers are equal. For example, $[9, 10]$, $[3, 3, 3]$, and $[9, 7, 5, 3]$ are arithmetic arrays, while $[1, 3, 3, 7]$, $[2, 1, 2]$, and $[1, 2, 4]$ are not.
Sarasvati again has an array of $N$ non-negative integers. The $i$-th integer of the array is $A_i$. She can replace at most $1$ element in the array with any (possibly negative) integer she wants.
For an array $A$, Sarasvati defines a subarray as any contiguous part of $A$. Please help Sarasvati determine the length of the longest possible arithmetic subarray she can create by replacing at most $1$ element in the original array.
Input Format
The first line of the input gives the number of test cases, $T$. $T$ test cases follow. Each test case begins with a line containing the integer $N$. The second line contains $N$ integers. The $i$-th integer is $A_i$.
Output Format
For each test case, output one line containing `Case #x:` followed by $y$, where $x$ is the test case number (starting from $1$) and $y$ is the length of the longest arithmetic subarray.
Explanation/Hint
In Sample Case #$1$, the whole array is an arithmetic array, thus the longest arithmetic subarray is the whole array.
In Sample Case #$2$, if Sarasvati changes the number at third position to $5$, the array will become $[5, 5, 5, 5, 5, 5, 4, 5, 6]$. The subarray from first position to sixth position is the longest arithmetic subarray.
In Sample Case #$3$, Sarasvati can change the number at the last position to $-1$, to get $[8, 5, 2, -1]$. This resulting array is arithmetic.
### Limits
$1 \le T \le 100$.
$0 \le A_i \le 10^9$.
**Test Set $1$**
$2 \le N \le 2000$.
**Test Set $2$**
$2 \le N \le 3 \times 10^5$ for at most $10$ test cases.
For the remaining cases, $2 \le N \le 2000$.