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 <=10^{5}

**Example:**

**Input:**

2

12

24

**Output:**

13

27

**Explanation:
Testcase1:**

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

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.

Author: sujnesh

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