#include <iostream.h>

// Problem 11480
// Let G,B,R be the number of balls of each colour.
// The range of values for B is between 2 and floor(N/2)-1  
// Imagine B is fixed.  Then R must satisfy 2 constraints:
// (1) R <= B-1
// (2) R+B+(B+1) <= R+B+G = N, which means R <= N-2B-1
// Thus, R can take any value between 1 and min(B-1,N-2B-1)
// Once B and R are chosen, G is uniquely determined.
// So, we just have to add up min(B-1,N-2B-1) for B=2..floor(N/2)-1
// Now, B-1 <= N-2B-1 iff B <= N/3
// So we can split up the sum into two pieces:
// The sum of (B-1) for B=2..floor(N/3), which is (floor(N/3)-1)floor(N/3)/2
// and the sum of (N-2B-1) for B=floor(N/3)+1 to floor(N/2)-1, which is
// (N-1)(floor(N/2)-1-floor(N/3))-(floor(N/2)-1)floor(N/2)
//    +floor(N/3)(floor(N/3)+1)

int main() {
  long long N;
  int counter = 1;
  cin >> N;
  while (N) {
    long long N3 = N/3;
    long long N2 = N/2;
    long long r = (N3-1)*N3/2+(N-1)*(N2-1-N3)-(N2-1)*N2+N3*(N3+1);
    cout << "Case " << counter << ": " << r << "\n";
    counter++;
    cin >> N;
  }
}