CF2107F1 Cycling (Easy Version)

This is the easy version of the problem. The difference between the versions is that in this version, $ 1\le n\le 5\cdot 10^3 $ and you don't need to output the answer for each prefix. You can hack only if you solved all versions of this problem. Leo works as a programmer in the city center, and his lover teaches at a high school in the suburbs. Every weekend, Leo would ride his bike to the suburbs to spend a nice weekend with his lover. There are $ n $ cyclists riding in front of Leo on this road right now. They are numbered $ 1 $ , $ 2 $ , $ \ldots $ , $ n $ from front to back. Initially, Leo is behind the $ n $ -th cyclist. The $ i $ -th cyclist has an agility value $ a_i $ . Leo wants to get ahead of the $ 1 $ -st cyclist. Leo can take the following actions as many times as he wants: - Assuming that the first person in front of Leo is cyclist $ i $ , he can go in front of cyclist $ i $ for a cost of $ a_i $ . This puts him behind cyclist $ i - 1 $ . - Using his super powers, swap $ a_i $ and $ a_j $ ( $ 1\le i < j\le n $ ) for a cost of $ (j - i) $ . Leo wants to know the minimum cost to get in front of the $ 1 $ -st cyclist. Here you only need to print the answer for the whole array, i.e. $ [a_1, a_2, \ldots, a_n] $ .

Each test contains multiple test cases. The first line contains the number of test cases $ t $ ( $ 1 \le t \le 10^3 $ ). The description of the test cases follows. The first line of each test case contains a positive integer $ n $ ( $ 1 \leq n \leq 5\cdot 10^3 $ ), representing the number of the cyclists. The second line of each test case contains $ n $ integers $ a_1, \ldots, a_n $ ( $ 1 \leq a_i \leq 10^9 $ ). It is guaranteed that the sum of $ n $ over all test cases does not exceed $ 5\cdot 10^3 $ .

For each test case, print one integer representing the minimum cost for Leo to go from behind the $ n $ -th cyclist to in front of the $ 1 $ -st cyclist.

In the first test case, one possible way to move from the position behind the $ n $ -th cyclist to the position in front of the $ 1 $ -st cyclist is: - Leo swaps $ a_2 $ $ (i=2) $ and $ a_3 $ $ (j=3) $ , then the array becomes $ [1,4,2] $ ; it costs $ j-i=3-2=1 $ . - Leo is behind the $ 3 $ -rd cyclist and moves behind the $ 2 $ -nd cyclist; it costs $ a_3=2 $ . - Leo swaps $ a_1 $ $ (i=1) $ and $ a_2 $ $ (j=2) $ , then the array becomes $ [4,1,2] $ ; it costs $ j-i=2-1=1 $ . - Leo is behind the $ 2 $ -nd cyclist and moves behind the $ 1 $ -st cyclist; it costs $ a_2=1 $ . - Leo swaps $ a_1 $ $ (i=1) $ and $ a_2 $ $ (j=2) $ , then the array becomes $ [1,4,2] $ ; it costs $ j-i=2-1=1 $ . - Leo moves ahead of the $ 1 $ -st cyclist; it costs $ a_1=1 $ . So the total cost is $ 1+2+1+1+1+1=7 $ . It can be proved that $ 7 $ is the minimum cost. In the second test case, to move ahead of the $ 1 $ -st cyclist from the position behind the $ n $ -th cyclist, Leo should not swap anyone's agility value. The total cost is $ 1+1+1+1=4 $ .