Chaotic V.

设$x$是大于$1$的正整数，定义$f(x)$为$x$的最小质因子的值。 现在考虑构造一棵树$T$。其中节点均用从$1$开始的正整数编号，对于任意大于$1$的正整数$x$，节点$x$在树上的父亲是$\frac{x}{f(x)}$。 在这棵树上有$n$个关键节点，很特别地，这$n$个节点的编号均可以表示成一个非负整数的阶乘。更具体地，第$i$个节点$P_i$的编号为$k_i!=\prod_{j=1}^{k_i}j$。 定义$\text{DIS}(i,j)$表示树上$i,j$两点间最短路所包含的边数。现在你找出树上的一个节点$P$，最小化$\sum_{i=1}^n\text{DIS}(P,P_i)$的值。你只需要输出这个最小值即可。 **输入格式** 第一行一个正整数$n(1\leq n\leq 10^6)$，代表节点数。 第二行$n$个非负整数$k_i(0\leq k_i\leq 5\times 10^3)$，描述节点$P_i$的编号为$k_i!$。 **输出格式** 一行一个非负整数表示答案。

[Æsir - CHAOS](https://soundcloud.com/kivawu/aesir-chaos) [Æsir - V.](https://soundcloud.com/kivawu/aesir-v) "Everything has been planned out. No more hidden concerns. The condition of Cytus is also perfect. The time right now...... 00:01:12...... It's time." The emotion samples are now sufficient. After almost 3 years, it's time for Ivy to awake her bonded sister, Vanessa. The system inside A.R.C.'s Library core can be considered as an undirected graph with infinite number of processing nodes, numbered with all positive integers ( $ 1, 2, 3, \ldots $ ). The node with a number $ x $ ( $ x > 1 $ ), is directly connected with a node with number $ \frac{x}{f(x)} $ , with $ f(x) $ being the lowest prime divisor of $ x $ . Vanessa's mind is divided into $ n $ fragments. Due to more than 500 years of coma, the fragments have been scattered: the $ i $ -th fragment is now located at the node with a number $ k_i! $ (a factorial of $ k_i $ ). To maximize the chance of successful awakening, Ivy decides to place the samples in a node $ P $ , so that the total length of paths from each fragment to $ P $ is smallest possible. If there are multiple fragments located at the same node, the path from that node to $ P $ needs to be counted multiple times. In the world of zeros and ones, such a requirement is very simple for Ivy. Not longer than a second later, she has already figured out such a node. But for a mere human like you, is this still possible? For simplicity, please answer the minimal sum of paths' lengths from every fragment to the emotion samples' assembly node $ P $ .

The first line contains an integer $ n $ ( $ 1 \le n \le 10^6 $ ) — number of fragments of Vanessa's mind. The second line contains $ n $ integers: $ k_1, k_2, \ldots, k_n $ ( $ 0 \le k_i \le 5000 $ ), denoting the nodes where fragments of Vanessa's mind are located: the $ i $ -th fragment is at the node with a number $ k_i! $ .

Print a single integer, denoting the minimal sum of path from every fragment to the node with the emotion samples (a.k.a. node $ P $ ). As a reminder, if there are multiple fragments at the same node, the distance from that node to $ P $ needs to be counted multiple times as well.

3
2 1 4

5

4
3 1 4 4

6

4
3 1 4 1

6

5
3 1 4 1 5

11

Considering the first $ 24 $ nodes of the system, the node network will look as follows (the nodes $ 1! $ , $ 2! $ , $ 3! $ , $ 4! $ are drawn bold):  For the first example, Ivy will place the emotion samples at the node $ 1 $ . From here: - The distance from Vanessa's first fragment to the node $ 1 $ is $ 1 $ . - The distance from Vanessa's second fragment to the node $ 1 $ is $ 0 $ . - The distance from Vanessa's third fragment to the node $ 1 $ is $ 4 $ . The total length is $ 5 $ . For the second example, the assembly node will be $ 6 $ . From here: - The distance from Vanessa's first fragment to the node $ 6 $ is $ 0 $ . - The distance from Vanessa's second fragment to the node $ 6 $ is $ 2 $ . - The distance from Vanessa's third fragment to the node $ 6 $ is $ 2 $ . - The distance from Vanessa's fourth fragment to the node $ 6 $ is again $ 2 $ . The total path length is $ 6 $ .