#CUBNUM. Cube Numbers

    ID: 3423 远端评测题 3000ms 1536MiB 尝试: 0 已通过: 0 难度: (无) 上传者: 标签>simple-mathdynamic-programming

Cube Numbers

本题没有可用的提交语言。

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