CF1092B.Teams Forming

There are $n$ students in a university. The number of students is even. The $i$-th student has programming skill equal to $a_i$.

The coach wants to form $\frac{n}{2}$ teams. Each team should consist of exactly two students, and each student should belong to exactly one team. Two students can form a team only if their skills are equal (otherwise they cannot understand each other and cannot form a team).

Students can solve problems to increase their skill. One solved problem increases the skill by one.

The coach wants to know the minimum total number of problems students should solve to form exactly $\frac{n}{2}$ teams (i.e. each pair of students should form a team). Your task is to find this number.

某所大学有 $n$ 名学生，学生人数为偶数。第 $i$ 名学生的编程能力为 $a_i$。

教练希望组成 $\frac{n}{2}$ 支队伍。每支队伍必须恰好由两名学生组成，且每名学生必须恰好属于一支队伍。两名学生仅当其能力值相等时才能组成一支队伍（否则他们无法相互理解，因而无法组队）。

学生可以通过解题来提升自己的能力。每解出一道题可使其能力值增加 $1$。

教练希望知道：为恰好组成 $\frac{n}{2}$ 支队伍（即所有学生两两配对成队），学生们需要解决的题目总数的最小值。你的任务就是求出这个最小值。

The first line of the input contains one integer $n$ ($2 \le n \le 100$) — the number of students. It is guaranteed that $n$ is even.

The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the skill of the $i$-th student.

输入的第一行包含一个整数 $n$（$2 \le n \le 100$）—— 学生人数。保证 $n$ 为偶数。

输入的第二行包含 $n$ 个整数 $a_1, a_2, \dots, a_n$（$1 \le a_i \le 100$），其中 $a_i$ 表示第 $i$ 个学生的技能值。

Print one number — the minimum total number of problems students should solve to form exactly $\frac{n}{2}$ teams.

输出一个数字——学生需要解决的最少问题总数，以恰好组成 $\frac{n}{2}$ 个队伍。

In the first example the optimal teams will be: $(3, 4)$, $(1, 6)$ and $(2, 5)$, where numbers in brackets are indices of students. Then, to form the first team the third student should solve $1$ problem, to form the second team nobody needs to solve problems and to form the third team the second student should solve $4$ problems so the answer is $1 + 4 = 5$.

In the second example the first student should solve $99$ problems to form a team with the second one.

在第一个例子中，最优的队伍为：$(3, 4)$、$(1, 6)$ 和 $(2, 5)$，其中括号内的数字表示学生的索引。那么，为组成第一支队伍，第三名学生需解决 $1$ 个问题；为组成第二支队伍，无需任何学生解决问题；为组成第三支队伍，第二名学生需解决 $4$ 个问题，因此答案为 $1 + 4 = 5$。

在第二个例子中，第一名学生需解决 $99$ 个问题才能与第二名学生组成一支队伍。