CF1728F.Fishermen

There are $n$ fishermen who have just returned from a fishing trip. The $i$-th fisherman has caught a fish of size $a_i$.

The fishermen will choose some order in which they are going to tell the size of the fish they caught (the order is just a permutation of size $n$). However, they are not entirely honest, and they may "increase" the size of the fish they have caught.

Formally, suppose the chosen order of the fishermen is $[p_1, p_2, p_3, \dots, p_n]$. Let $b_i$ be the value which the $i$-th fisherman in the order will tell to the other fishermen. The values $b_i$ are chosen as follows:

• the first fisherman in the order just honestly tells the actual size of the fish he has caught, so $b_1 = a_{p_1}$;
• every other fisherman wants to tell a value that is strictly greater than the value told by the previous fisherman, and is divisible by the size of the fish that the fisherman has caught. So, for $i \gt 1$, $b_i$ is the smallest integer that is both strictly greater than $b_{i-1}$ and divisible by $a_{p_i}$.

For example, let $n = 7$, $a = [1, 8, 2, 3, 2, 2, 3]$. If the chosen order is $p = [1, 6, 7, 5, 3, 2, 4]$, then:

• $b_1 = a_{p_1} = 1$;
• $b_2$ is the smallest integer divisible by $2$ and greater than $1$, which is $2$;
• $b_3$ is the smallest integer divisible by $3$ and greater than $2$, which is $3$;
• $b_4$ is the smallest integer divisible by $2$ and greater than $3$, which is $4$;
• $b_5$ is the smallest integer divisible by $2$ and greater than $4$, which is $6$;
• $b_6$ is the smallest integer divisible by $8$ and greater than $6$, which is $8$;
• $b_7$ is the smallest integer divisible by $3$ and greater than $8$, which is $9$.

You have to choose the order of fishermen in a way that yields the minimum possible $\sum\limits_{i=1}^{n} b_i$.

有 $n$ 位渔民刚刚结束垂钓返回。第 $i$ 位渔民捕获了一条大小为 $a_i$ 的鱼。

这些渔民将选择一个顺序（即 $n$ 个渔民的一个排列）来依次报出自己所捕获鱼的大小。但他们并非完全诚实，可能会“夸大”自己所捕获鱼的大小。

形式化地，假设选定的渔民顺序为 $[p_1, p_2, p_3, \dots, p_n]$。令 $b_i$ 表示该顺序中第 $i$ 位渔民向其他人报出的数值。这些值 $b_i$ 按如下规则确定：

• 顺序中第一位渔民如实报告自己所捕获鱼的实际大小，即 $b_1 = a_{p_1}$；
• 其余每位渔民都希望报出一个严格大于前一位渔民所报数值、且能被自己所捕获鱼的大小整除的数。因此，对 $i > 1$，$b_i$ 是满足以下两个条件的最小整数：(1) 严格大于 $b_{i-1}$；(2) 被 $a_{p_i}$ 整除。

例如，设 $n = 7$，$a = [1, 8, 2, 3, 2, 2, 3]$。若选定顺序为 $p = [1, 6, 7, 5, 3, 2, 4]$，则：

• $b_1 = a_{p_1} = 1$；
• $b_2$ 是大于 $1$ 且能被 $2$ 整除的最小整数，即 $2$；
• $b_3$ 是大于 $2$ 且能被 $3$ 整除的最小整数，即 $3$；
• $b_4$ 是大于 $3$ 且能被 $2$ 整除的最小整数，即 $4$；
• $b_5$ 是大于 $4$ 且能被 $2$ 整除的最小整数，即 $6$；
• $b_6$ 是大于 $6$ 且能被 $8$ 整除的最小整数，即 $8$；
• $b_7$ 是大于 $8$ 且能被 $3$ 整除的最小整数，即 $9$。

你需要选择一个渔民顺序，使得 $\sum\limits_{i=1}^{n} b_i$ 的值尽可能小。

The first line contains one integer $n$ ($1 \le n \le 1000$) — the number of fishermen.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$).

第一行包含一个整数 $n$（$1 \le n \le 1000$）—— 渔民的数量。

第二行包含 $n$ 个整数 $a_1, a_2, \dots, a_n$（$1 \le a_i \le 10^6$）。

Print one integer — the minimum possible value of $\sum\limits_{i=1}^{n} b_i$ you can obtain by choosing the order of fishermen optimally.

输出一个整数——通过最优地选择渔民的顺序，所能得到的 $\sum\limits_{i=1}^{n} b_i$ 的最小可能值。