Toilet Paper Mysteries Today you need to solve a very practical problem: How many loops do you need to have a certain number of sheets on your toilet paper roll? Let's look at some facts:


The diameter of a bare toilet paper cylinder is 3.8cm
The length of one sheet of toilet paper is 10cm.
The thickness of one sheet of toilet paper is 1mm. 


Before you wrap around the cylinder the first time, it has a circumference in cm of 3.8*pi. Every time you wrap a sheet around the cylinder its radius increases by .1, therefore its circumference increases by .2*PI. Use this information to find out how many loops it takes to fit n sheets of toilet paper. (Note: Use an approximation of Pi that is at least as accurate as 3.14159).

Test Cases: 

n=1:


10/(3.8*pi) = .838 loops


n=2:


(How many full loops can we make?) 1 full loop = 3.8*pi = 11.938.
(How much do we have left after the 1st loop?) 20 - 11.938 = 8.062
(How much of a 2nd loop does the remaining piece make?) 8.062/(4*pi) = .642 loops
Answer: 1.642 loops


n=3:


1st full loop = 3.8*pi = 11.938, 2nd full loop = 4*pi = 12.566
30 - 11.938 - 12.566 = 5.496
5.496/(4.2*pi) = .417
Answer: 2.417 loops


n=100 => 40.874
Pyth, 27 23 bytes

+fg0=-QJc*.n0+18T50)cQJ


Try it online. Test suite.


Explanation

                            Q = input number (implicit)
 f                 )        increment T from 1, for each T:
             +18T             add 18 to T, get radius
         *.n0                 multiply by pi to, get half the circumference
        c        50           divide by 50, get circumference in sheets
       J                      save it to J
    =-Q                       decrement Q by it
  g0                          use this T if Q is now &lt;= 0
+                           add
                     Q        Q (now &lt;= 0)
                    c J       divided by J (the last circumference)
                            and print (implicit)