C. The Sports Festivaltime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputThe student council is preparing for the relay race at the sports festival.The council consists of n members. They will run one after the other in the race, the speed of member i is s_i. The discrepancy d_i of the i-th stage is the difference between the maximum and the minimum running speed among the first i members who ran. Formally, if a_i denotes the speed of the i-th member who participated in the race, then d_i = \max(a_1, a_2, \dots, a_i) - \min(a_1, a_2, \dots, a_i).You want to minimize the sum of the discrepancies d_1 + d_2 + \dots + d_n. To do this, you are allowed to change the order in which the members run. What is the minimum possible sum that can be achieved?InputThe first line contains a single integer n (1 \le n \le 2000)  — the number of members of the student council.The second line contains n integers s_1, s_2, \dots, s_n (1 \le s_i \le 10^9)  – the running speeds of the members.OutputPrint a single integer  — the minimum possible value of d_1 + d_2 + \dots + d_n after choosing the order of the members.ExamplesInput
3
3 1 2
Output
3
Input
1
5
Output
0
Input
6
1 6 3 3 6 3
Output
11
Input
6
104 943872923 6589 889921234 1000000000 69
Output
2833800505
NoteIn the first test case, we may choose to make the third member run first, followed by the first member, and finally the second. Thus a_1 = 2, a_2 = 3, and a_3 = 1. We have:  d_1 = \max(2) - \min(2) = 2 - 2 = 0.  d_2 = \max(2, 3) - \min(2, 3) = 3 - 2 = 1.  d_3 = \max(2, 3, 1) - \min(2, 3, 1) = 3 - 1 = 2. The resulting sum is d_1 + d_2 + d_3 = 0 + 1 + 2 = 3. It can be shown that it is impossible to achieve a smaller value.In the second test case, the only possible rearrangement gives d_1 = 0, so the minimum possible result is 0.
3
3 1 2

3 
['dp', 'greedy', '*1800']