#P1859C. Another Permutation Problem
Another Permutation Problem
No submission language available for this problem.
Description
Andrey is just starting to come up with problems, and it's difficult for him. That's why he came up with a strange problem about permutations$^{\dagger}$ and asks you to solve it. Can you do it?
Let's call the cost of a permutation $p$ of length $n$ the value of the expression:
Find the maximum cost among all permutations of length $n$.
$^{\dagger}$A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$ but there is $4$ in the array).
Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 30$) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer $n$ ($2 \le n \le 250$) — the length of the permutation.
It is guaranteed that the sum of $n$ over all test cases does not exceed $500$.
For each test case, output a single integer — the maximum cost among all permutations of length $n$.
Input
Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 30$) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer $n$ ($2 \le n \le 250$) — the length of the permutation.
It is guaranteed that the sum of $n$ over all test cases does not exceed $500$.
Output
For each test case, output a single integer — the maximum cost among all permutations of length $n$.
5
2
4
3
10
20
2
17
7
303
2529
Note
In the first test case, the permutation with the maximum cost is $[2, 1]$. The cost is equal to $2 \cdot 1 + 1 \cdot 2 - \max (2 \cdot 1, 1 \cdot 2)= 2 + 2 - 2 = 2$.
In the second test case, the permutation with the maximum cost is $[1, 2, 4, 3]$. The cost is equal to $1 \cdot 1 + 2 \cdot 2 + 4 \cdot 3 + 3 \cdot 4 - 4 \cdot 3 = 17$.