National Olympiad in Informatics, China, 2001

The inverse tangent function can be expressed as an infinite series, as shown below:

$$\displaystyle \begin{align} \arctan(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1} \quad (0 \le x \le 1) \end{align} \tag{1}$$

It is commonly known that the inverse tangent function can be used to compute ~\pi~. For example, an easy way to compute ~\pi~ is using the method:

$$\displaystyle \begin{align} \pi &= 4\arctan(1) \\ &= 4 \left(1-\frac 1 3+\frac 1 5-\frac 1 7+\frac 1 9-\frac 1 {11}+\dots\right) \end{align} \tag{2}$$

Of course, this method is rather inefficient. We can apply the tangent angle sum identity:

$$\displaystyle \begin{align} \tan(\alpha+\beta) = \frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)} \end{align} \tag{3}$$

After some simple manipulation, the following is obtained:

$$\displaystyle \begin{align} \arctan(p) + \arctan(q) = \arctan\left(\frac{p+q}{1-pq}\right) \end{align} \tag{4}$$

Using this identity, let ~p=\dfrac{1}{2}~ and ~q=\dfrac{1}{3}~, then ~\dfrac{p+q}{1-pq}=1~. Therefore:

$$\displaystyle \arctan\left(\frac 1 2\right)+\arctan\left(\frac 1 3\right) = \arctan\left(\frac{\frac 1 2+\frac 1 3}{1-\frac 1 2 \cdot \frac 1 3}\right) = \arctan(1)$$

Using the inverse tangents of ~\dfrac 1 2~ and ~\dfrac 1 3~ to calculate ~\arctan(1)~, the speed is drastically improved.

We take equation (4) and write it in the following form:

$$\displaystyle \arctan\left(\frac 1 a\right) = \arctan\left(\frac 1 b\right)+\arctan\left(\frac 1 c\right)$$

where ~a~, ~b~, and ~c~ are each positive integers.

The problem is, for a given ~a~ ~(1 \le a \le 60\,000)~, find the value of ~b + c~. It is guaranteed that for any ~a~ there will always exist an integer solution. If there are multiple solutions, you are required to find the minimum value of ~b + c~.

Input Specification

The input consists of a single positive integer ~a~, where ~1 \le a \le 60\,000~.

Output Specification

The output should contain a single integer, the value of ~b + c~.

Sample Input

1

Sample Output

5

Problem translated to English by Alex.