#P1859C. Another Permutation Problem
Another Permutation 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 $1 \cdot 1 + 2 \cdot 2 + 4 \cdot 3 + 3 \cdot 4 - 4 \cdot 3 = 17$.