Sasha and Magnetic Machines

题意翻译

### 题意简述 有一个长度为$n$的正整数序列$a_{1..n}$。你可以对这个数列进行最多$1$次的如下操作: - 选择$1 \leq i,j \leq n\ \ (i \neq j)$,并选择一个可以整除$a_i$的正整数$x$,然后将$a_i$变为$\frac{a_i}{x}$,将$a_j$变为$a_j \cdot x$。 问你操作后,该序列中所有数的和最小能达到多少。 ### 输入格式 第一行输入$n\ (2\leq n\leq 5\cdot10^4)$,表示序列长度。 第二行输入$n$个空格隔开的正整数$a_i\ (1 \leq a_i \leq 100)$,描述序列中的元素。 ### 输出格式 输出一个整数,操作后序列最小可能的数值和。

题目描述

One day Sasha visited the farmer 2D and his famous magnetic farm. On this farm, the crop grows due to the influence of a special magnetic field. Maintaining of the magnetic field is provided by $ n $ machines, and the power of the $ i $ -th machine is $ a_i $ . This year 2D decided to cultivate a new culture, but what exactly he didn't say. For the successful growth of the new culture, it is necessary to slightly change the powers of the machines. 2D can at most once choose an arbitrary integer $ x $ , then choose one machine and reduce the power of its machine by $ x $ times, and at the same time increase the power of one another machine by $ x $ times (powers of all the machines must stay positive integers). Note that he may not do that if he wants. More formally, 2D can choose two such indices $ i $ and $ j $ , and one integer $ x $ such that $ x $ is a divisor of $ a_i $ , and change powers as following: $ a_i = \frac{a_i}{x} $ , $ a_j = a_j \cdot x $ Sasha is very curious, that's why he wants to calculate the minimum total power the farmer can reach. There are too many machines, and Sasha can't cope with computations, help him!

输入输出格式

输入格式


The first line contains one integer $ n $ ( $ 2 \le n \le 5 \cdot 10^4 $ ) — the number of machines. The second line contains $ n $ integers $ a_1, a_2, \ldots, a_n $ ( $ 1 \le a_i \le 100 $ ) — the powers of the machines.

输出格式


Print one integer — minimum total power.

输入输出样例

输入样例 #1

5
1 2 3 4 5

输出样例 #1

14

输入样例 #2

4
4 2 4 4

输出样例 #2

14

输入样例 #3

5
2 4 2 3 7

输出样例 #3

18

说明

In the first example, the farmer can reduce the power of the $ 4 $ -th machine by $ 2 $ times, and increase the power of the $ 1 $ -st machine by $ 2 $ times, then the powers will be: $ [2, 2, 3, 2, 5] $ . In the second example, the farmer can reduce the power of the $ 3 $ -rd machine by $ 2 $ times, and increase the power of the $ 2 $ -nd machine by $ 2 $ times. At the same time, the farmer can leave is be as it is and the total power won't change. In the third example, it is optimal to leave it be as it is.