Description: The 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? Input Format: The 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. Output Format: Print a single integer — the minimum possible value of $$$d_1 + d_2 + \dots + d_n$$$ after choosing the order of the members. Note: In 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 $$$0$$$.