Given n, there can be n! circular arrangement of the numbers 0 to n − 1. Lets represent every permutation as P1 P2 P3 . . . Pn! SOP(Pk) = sum of product of every two contiguous numbers in Pk. Consider an example where n = 4 and Pk = (1 3 2 0), therefore SOP(Pk) = 1∗3+3∗2+2∗0+0∗1 = 9. You have to find out the number of distinct values of SOP(Pk) for k = 1 to n!. For n = 3, Pk Permutation SOP(Pk) P1 012 2 P2 021 2 P3 102 2 P4 120 2 P5 201 2 P6 210 2 So, for n = 3, there is only 1 distinct value of SOP(Pk). Input There will be multiple test cases. Each case consists of a line containing a positive integer n (1 < n ≤ 20). The last line of input file contains a single ‘0’ that doesn’t need to be processed. The total number of test cases will be at most 30. Output For each case, output the case number followed by the number of distinct SOPs. Sample Input 3 4 6 0 Sample Output Case #1: 1 Case #2: 3 Case #3: 21