Description

In mathematics, the n^th harmonic number is the sum of the reciprocals of the first n natural numbers:

In this problem, you are given n, you have to find H_n.

Input

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

Each case starts with a line containing an integer n (1 ≤ n ≤ 10⁸

Output

For each case, print the case number and the n^th harmonic number. Errors less than 10^-8 will be ignored.

Sample Input

12

1

2

3

4

5

6

7

8

9

90000000

99999999

100000000

Sample Output

Case 1: 1

Case 2: 1.5

Case 3: 1.8333333333

Case 4: 2.0833333333

Case 5: 2.2833333333

Case 6: 2.450

Case 7: 2.5928571429

Case 8: 2.7178571429

Case 9: 2.8289682540

Case 10: 18.8925358988

Case 11: 18.9978964039

Case 12: 18.9978964139

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<ctype.h>
#include<math.h>
#include<iostream>
 

using namespace std;
#include<algorithm>
#define N 2100
double a[10010];
#define Max 0x3f3f3f3f
int main()
{
    int i, n, T, t=1;
    scanf("%d", &T);
    while(T--)
    {
        a[1]=1;
        for(i=2;i<10000;i++)
        {
            a[i]=a[i-1]+1.0/i;
        }
        scanf("%d", &n);
        printf("Case %d: ", t++);
        if(n<10000)
        {
            /*if(n==1)
                printf("1\n");
            else if(n==2)
                printf("1.5\n");
            else if(n==6)
                printf("2.450\n");
            else*/
                printf("%.10f\n", a[n]);
        }
        else
        {
            double sum;
            sum=log(n)+0.5772156649+1.0/(2*n);
            printf("%.10f\n", sum);
        }
    }
    return 0;
}