Codeforces Round 633 (Div. 2)
B
Sorted Adjacent Differences
Codeforces
1200
原题
← AC →
翻译中…

You have array of $n$ numbers $a_{1}, a_{2}, \ldots, a_{n}$.

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Rearrange these numbers to satisfy $|a_{1} - a_{2}| \le |a_{2} - a_{3}| \le \ldots \le |a_{n-1} - a_{n}|$, where $|x|$ denotes absolute value of $x$. It's always possible to find such rearrangement.

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Note that all numbers in $a$ are not necessarily different. In other words, some numbers of $a$ may be same.

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You have to answer independent $t$ test cases.

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Input

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The first line contains a single integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases.

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The first line of each test case contains single integer $n$ ($3 \le n \le 10^{5}$) — the length of array $a$. It is guaranteed that the sum of values of $n$ over all test cases in the input does not exceed $10^{5}$.

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The second line of each test case contains $n$ integers $a_{1}, a_{2}, \ldots, a_{n}$ ($-10^{9} \le a_{i} \le 10^{9}$).

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Output

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For each test case, print the rearranged version of array $a$ which satisfies given condition. If there are multiple valid rearrangements, print any of them.

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Note

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In the first test case, after given rearrangement, $|a_{1} - a_{2}| = 0 \le |a_{2} - a_{3}| = 1 \le |a_{3} - a_{4}| = 2 \le |a_{4} - a_{5}| = 2 \le |a_{5} - a_{6}| = 10$. There are other possible answers like "5 4 5 6 -2 8".

翻译中…

In the second test case, after given rearrangement, $|a_{1} - a_{2}| = 1 \le |a_{2} - a_{3}| = 2 \le |a_{3} - a_{4}| = 4$. There are other possible answers like "2 4 8 1".

翻译中…

样例

Input #1
2
6
5 -2 4 8 6 5
4
8 1 4 2
Output #1
5 5 4 6 8 -2
1 2 4 8
InputCopy OutputCopy
题目信息
题号
1339B
时间限制
1 s
内存限制
256 MB
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