F. Fishermentime limit per test6 secondsmemory limit per test512 megabytesinputstandard inputoutputstandard outputThere 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 > 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.InputThe 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).OutputPrint one integer — the minimum possible value of \sum\limits_{i=1}^{n} b_i you can obtain by choosing the order of fishermen optimally.ExamplesInput
7
1 8 2 3 2 2 3
Output
33
Input
10
5 6 5 6 5 6 5 6 5 6
Output
165

7
1 8 2 3 2 2 3

33 
['flows', 'graph matchings', 'greedy', '*3100']