Rimi learned a new thing about integers, which is - any positive integer greater than 1 can be divided by its divisors. So, he is now playing with this property. He selects a number N. And he calls this D.

In each turn he randomly chooses a divisor of D (1 to D). Then he divides D by the number to obtain new D

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case begins with an integer N (1 ≤ N ≤ 105).

Output

For each case of input you have to print the case number and the expected value. Errors less than 10-6

Sample Input

3

1

2

50

Sample Output

Case 1: 0

Case 2: 2.00

Case 3: 3.0333333333

#include<bits/stdc++.h>
#define ll long long
using namespace std;
double dp[100010];
void init()
{
    dp[1]=0;
    for(int i=2;i<=100000;i++)
    {
        int cnt=0;
        double ans=0.0;
        for(int j=1;j*j<=i;j++)
        {
            if(i%j==0)
            {
                cnt++;
                ans+=dp[j];
                if(i/j!=j)
                {
                    cnt++;
                    ans+=dp[i/j];
                }
            }
        }
        dp[i]=(ans/(cnt-1))+((cnt)*1.0/(cnt-1));
    }
}
int main()
{
    init();
    int t,n;
    scanf("%d",&t);
   for(int oo=1;oo<=t;oo++)
    {
        scanf("%d",&n);
        printf("Case %d: %lf\n",oo,dp[n]);
    }
}