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Maximum sum Problem
##### Submissions: 13282   Accuracy: 50.9%   Difficulty: Easy   Marks: 2

Given a number n, we can divide it in only three parts n/2, n/3 and n/4 (we will consider only integer part). The task is to find the maximum sum we can make by dividing number in three parts recursively and summing up them together.
Note: Sometimes, the maximum sum can be obtained by not dividing n.

Input:
The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. The first line of each test case contains the integer n.

Output:
For each testcase, in a new line, print the maximum sum possible.

Constraints:
1<= T <=100
1<= n <=105

Example:
Input:
2
12
24

Output:
13
27

Explanation:
Testcase1:

Input : n = 12
Output : 13
We break n = 12 in three parts {12/2, 12/3, 12/4} = {6, 4, 3},  now current sum is = (6 + 4 + 3) = 13
Again, we break 6 = {6/2, 6/3, 6/4} = {3, 2, 1} = 3 + 2 + 1 = 6 and further breaking 3, 2 and 1 we get maximum
summation as 1, so breaking 6 in three parts produces maximum sum 6 only similarly breaking 4 in three parts we can get maximum sum 4 and same for 3 also. Thus maximum sum by breaking number in parts  is=13

Testcase2:
Input : n = 24
Output : 27
We break n = 24 in three parts {24/2, 24/3, 24/4} = {12, 8, 6},  now current sum is = (12 + 8 + 6) = 16
As seen in example, recursively breaking 12 would produce value 13. So our maximum sum is 13 + 8 + 6 = 27.
Note that recursively breaking 8 and 6 doesn't produce more values, that is why they are not broken further.

#### ** For More Input/Output Examples Use 'Expected Output' option **

Contributor: Sujnesh Mishra
Author: sujnesh

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