Harmonic Number 调和级数的欧拉公式

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

题意：

给定n,求调和级数Hn；

注意：

输出五花八门，关键是保留十位小数，直接输出%.10f就行

n<=1e8,直接开数组打表会ML,这时候要用欧拉公式,但是n很大时公式才成立，所以一部分打表，另一部分直接算，数组开到1e4就行。

Hn=ln(n)+c+1.0/(2*n)

c=0.57721566490153286060651209;

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