#CUBNUM. Cube Numbers

Cube Numbers

For any positive integer n, n can be represented as sum of other positive cube numbers (n = a13 + a23 + ... + am3). Your task is to print the smallest m, where m is number of cube numbers used to form n, such that n = a13 + a23 + ... + am3. For example:

  • n = 5, n = 13 + 13 + 13 + 13 + 13 (m = 5)
  • n = 8, n = 23 (m = 1)
  • n = 35, n = 23 + 33 (m = 2)

Note: My fastest time is 0.09s :).
Edit: My fastest time is 0.05s now lol
My Java solution is also accepted.

Input

Input consists of several test cases separated by new lines. Each test case consists of a positive integer, denoting the number of n (1 ≤ n ≤ 105). Input is terminated by end of file (EOF).
It is guaranteed that total test case per input file is less than 105.

Note: For c++ users, you can use while(scanf("%d",&n)!=EOF); to read input until EOF.
Warning: large Input/Output data, be careful with certain languages!.

Output

For each case, print "Case #X: M", where X (1 ≤ X ≤ 105) is the case number, and M is the minimum cube numbers used to form the integer n. There must be no trailing spaces at the end of printed lines, neither empty characters. Print a newline after each testcase.

Example

Input:
1
2
5
8
35

Output: Case #1: 1 Case #2: 2 Case #3: 5 Case #4: 1 Case #5: 2

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