Mad scientist Mike is building a time machine in his spare time. To finish the work, he needs a resistor with a certain resistance value.
However, all Mike has is lots of identical resistors with unit resistance
R0=1. Elements with other resistance can be constructed from these resistors. In this problem, we will consider the following as elements:
-
one resistor;
-
an element and one resistor plugged in sequence;
-
an element and one resistor plugged in parallel.
With the consecutive connection the resistance of the new element equals
R=Re+R0. With the parallel connection the resistance of the new element equals
. In this case
Re equals the resistance of the element being connected.
Mike needs to assemble an element with a resistance equal to the fraction
. Determine the smallest possible number of resistors he needs to make such an element.
Output
Print a single number − the answer to the problem.
Please do not use the
%lld specifier to read or write 64-bit integers in C++. It is recommended to use the
cin,
cout streams or the
%I64d specifier.
Note
In the first sample, one resistor is enough.
In the second sample one can connect the resistors in parallel, take the resulting element and connect it to a third resistor consecutively. Then, we get an element with resistance
. We cannot make this element using two resistors.