CF1106C.Lunar New Year and Number Division

Lunar New Year is approaching, and Bob is struggling with his homework – a number division problem.

There are $n$ positive integers $a_1, a_2, \ldots, a_n$ on Bob's homework paper, where $n$ is always an even number. Bob is asked to divide those numbers into groups, where each group must contain at least $2$ numbers. Suppose the numbers are divided into $m$ groups, and the sum of the numbers in the $j$ -th group is $s_j$ . Bob's aim is to minimize the sum of the square of $s_j$ , that is $\sum_{j = 1}^{m} s_j^2.$

Bob is puzzled by this hard problem. Could you please help him solve it?

The first line contains an even integer $n$ ( $2 \leq n \leq 3 \cdot 10^5$ ), denoting that there are $n$ integers on Bob's homework paper.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ( $1 \leq a_i \leq 10^4$ ), describing the numbers you need to deal with.

A single line containing one integer, denoting the minimum of the sum of the square of $s_j$ , which is $$$$\sum_{i = j}^{m} s_j^2, $$ where $m$$$ is the number of groups.

In the first sample, one of the optimal solutions is to divide those $4$ numbers into $2$ groups $\{2, 8\}, \{5, 3\}$ . Thus the answer is $(2 + 8)^2 + (5 + 3)^2 = 164$ .

In the second sample, one of the optimal solutions is to divide those $6$ numbers into $3$ groups $\{1, 2\}, \{1, 2\}, \{1, 2\}$ . Thus the answer is $(1 + 2)^2 + (1 + 2)^2 + (1 + 2)^2 = 27$ .