F1. Cycling (Easy Version)time limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputThis 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. $$$[a_1, a_2, \ldots, a_n]$$$.InputEach 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 $$$5\cdot 10^3$$$.OutputFor 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.ExampleInput431 2 441 1 1 121 244 1 3 2Output7 4 3 8 NoteIn 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 $$$1+1+1+1=4$$$.