Codeforces Round 815 (Div. 2)
B
Interesting Sum
Codeforces
800
原题
← AC →
翻译中…

You are given an array $a$ that contains $n$ integers. You can choose any proper subsegment $a_l, a_{l + 1}, \ldots, a_r$ of this array, meaning you can choose any two integers $1 \le l \le r \le n$, where $r - l + 1 \lt n$. We define the beauty of a given subsegment as the value of the following expression:

翻译中…

$$\max(a_{1}, a_{2}, \ldots, a_{l-1}, a_{r+1}, a_{r+2}, \ldots, a_{n}) - \min(a_{1}, a_{2}, \ldots, a_{l-1}, a_{r+1}, a_{r+2}, \ldots, a_{n}) + \max(a_{l}, \ldots, a_{r}) - \min(a_{l}, \ldots, a_{r}).$$

翻译中…

Please find the maximum beauty among all proper subsegments.

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Input

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The first line contains one integer $t$ ($1 \leq t \leq 1000$) — the number of test cases. Then follow the descriptions of each test case.

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The first line of each test case contains a single integer $n$ $(4 \leq n \leq 10^5)$ — the length of the array.

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The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_{i} \leq 10^9$) — the elements of the given array.

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It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

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Output

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For each testcase print a single integer — the maximum beauty of a proper subsegment.

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Note

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In the first test case, the optimal segment is $l = 7$, $r = 8$. The beauty of this segment equals to $(6 - 1) + (5 - 1) = 9$.

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In the second test case, the optimal segment is $l = 2$, $r = 4$. The beauty of this segment equals $(100 - 2) + (200 - 1) = 297$.

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样例

Input #1
4
8
1 2 2 3 1 5 6 1
5
1 2 3 100 200
4
3 3 3 3
6
7 8 3 1 1 8
Output #1
9
297
0
14
InputCopy OutputCopy
题目信息
题号
1720B
时间限制
1 s
内存限制
256 MB
标签
暴力
数据结构
贪心
数学
排序

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