Background

Little W is attending an arithmetic class.

Problem Description

As we all know, Gauss, when he was in elementary school, one day his math teacher assigned a problem: $1+2+\cdots+100=?$. The teacher thought it would take the students a long time to add 100 numbers, so he planned to sneak off for a coffee break. However, Gauss raised his hand and claimed he had finished the calculation—he used the well-known formula for the sum of an arithmetic series.

However, the teacher of this arithmetic class is as irresponsible as Gauss's teacher. He left an arithmetic problem on the blackboard and went off to chase after girls. But Little W is not as smart as Gauss, and this expression doesn't seem to have any clever shortcuts. Therefore, he contacted you, a student of competitive programming, via phone, hoping you could help him. In return, he will give you 100 points as a reward.

Specifically, the expression on the blackboard is as follows:

Let the prime factorization of $n$ be $n=\prod\limits_{i=1}^kp_i^{\alpha_i}$, then define $f(n)=\prod\limits_{i=1}^k{\dfrac{p_i^{\alpha_i+1}-1}{p_i^{\alpha_i+1}-p_i^{\alpha_i}}}$. Calculate the value of $\sum\limits_{i=1}^nf(i)$ (see Notes/Hints for precision requirements).

Note: Specifically, we define $f(1)=1$.

Input Format

A single line containing an integer $n$, as described above.

Output Format

A single line containing a real number representing the result.

Sample Input and Output

Sample Input #1

2

Sample Output #1

2.5000000000

Sample Input #2

5

Sample Output #2

6.7833333333

Notes/Hints

Explanation of Sample 2: $f(1)=1, f(2)=\dfrac{2^2-1}{2^2-2^1}=1.5, f(3)=\dfrac{3^2-1}{3^2-3^1}=1.3333333333, f(4)=\dfrac{2^3-1}{2^3-2^2}=1.75, f(5)=\dfrac{5^2-1}{5^2-5^1}=1.2$.

This problem includes a Special Judge (SPJ). Let your answer be $a$ and the standard answer be $b$.
If $|{a-b}|\le \min(\dfrac b{10^4},10)$, you will get all points for this test case;
Otherwise, if $|{a-b}|\le \min(\dfrac b{10^3},100)$, you will get 40% of the points for this test case;
Otherwise, you will get no points.
The standard answer will be rounded to 10 decimal places.
Note: Although this problem has lower precision requirements, it is still recommended to use long double for more accurate storage of the answer.

Data Range:
For 10% of the data, $n\le10$.
For 30% of the data, $n\le10^3$.
For 60% of the data, $n\le10^7$.
For 100% of the data, $1\le n\le10^{13}$.

见题面 Input 部分

见题面 Output 部分