# [Solution] Mainak and Array solution codeforces

Mainak and Array solution codeforces – Mainak has an array 𝑎1,𝑎2,,𝑎𝑛a1,a2,…,an of 𝑛n positive integers. He will do the following operation to this array exactly once:

• Pick a subsegment of this array and cyclically rotate it by any amount.

## [Solution] Mainak and Array solution codeforces

Formally, he can do the following exactly once:

• Pick two integers 𝑙l and 𝑟r, such that 1𝑙𝑟𝑛1≤l≤r≤n, and any positive integer 𝑘k.
• Repeat this 𝑘k times: set 𝑎𝑙=𝑎𝑙+1,𝑎𝑙+1=𝑎𝑙+2,,𝑎𝑟1=𝑎𝑟,𝑎𝑟=𝑎𝑙al=al+1,al+1=al+2,…,ar−1=ar,ar=al (all changes happen at the same time).

Mainak wants to maximize the value of (𝑎𝑛𝑎1)(an−a1) after exactly one such operation. Determine the maximum value of (𝑎𝑛𝑎1)(an−a1) that he can obtain.

Input

Each test contains multiple test cases. The first line contains a single integer 𝑡t (1𝑡501≤t≤50) — the number of test cases. Description of the test cases follows.

The first line of each test case contains a single integer 𝑛n (1𝑛20001≤n≤2000).

The second line of each test case contains 𝑛n integers 𝑎1,𝑎2,,𝑎𝑛a1,a2,…,an (1𝑎𝑖9991≤ai≤999).

It is guaranteed that the sum of 𝑛n over all test cases does not exceed 20002000.

## [Solution] Mainak and Array solution codeforces

For each test case, output a single integer — the maximum value of (𝑎𝑛𝑎1)(an−a1) that Mainak can obtain by doing the operation exactly once.

Example
input

Copy
5
6
1 3 9 11 5 7
1
20
3
9 99 999
4
2 1 8 1
3
2 1 5
output

Copy
10
0
990
7
4


## [Solution] Mainak and Array solution codeforces

• In the first test case, we can rotate the subarray from index 33 to index 66 by an amount of 22 (i.e. choose 𝑙=3l=3𝑟=6r=6 and 𝑘=2k=2) to get the optimal array:
[1,3,9,11,5,7⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯][1,3,5,7,9,11⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯][1,3,9,11,5,7_]⟶[1,3,5,7,9,11_]

So the answer is 𝑎𝑛𝑎1=111=10an−a1=11−1=10.

• In the second testcase, it is optimal to rotate the subarray starting and ending at index 11 and rotating it by an amount of 22.
• In the fourth testcase, it is optimal to rotate the subarray starting from index 11 to index 44 and rotating it by an amount of 11. So the answer is 81=78−1=7.