Problem Description

There is a youngster known for amateur propositions concerning several mathematical hard problems.

Nowadays, he is preparing a thought-provoking problem on a specific type of supercomputer which has ability to support calculations of integers between 0 and (2m−1) (inclusive).

As a young man born with ten fingers, he loves the powers of 10 so much, which results in his eccentricity that he always ranges integers he would like to use from 1 to 10k (inclusive).

For the sake of processing, all integers he would use possibly in this interesting problem ought to be as computable as this supercomputer could.

Given the positive integer m, your task is to determine maximum possible integer k that is suitable for the specific supercomputer.

Input

The input contains multiple test cases. Each test case in one line contains only one positive integer m, satisfying 1≤m≤105.

Output

For each test case, output “Case #x: y” in one line (without quotes), where x indicates the case number starting from 1 and y denotes the answer of corresponding case.

Sample Input

1 
64

Sample Output

Case #1: 0 
Case #2: 19

题目大意：

求输入一个数n，求2的n次方的位数-1，十进制中位数及10的次方数-1，所以该题公式为，m=int(n*log10(2))，m即为位数-1。

c++

#include<iostream> #include<cmath> using namespace std; int main() { long long int a,b,c,d; a=0; while(cin>>b) { a++; c=int(b*log10(2)); cout<<"Case #"<<a<<": "; cout<<c<<endl; } return 0; }