Nicola discovered a caliper inside a set of drafting tools he received as a gift. Seeing the caliper, he has decided to learn how to use it.

Through any three points that do not exist on the same line, there lies a unique circle. The points of this circle are represented in a string with the coordinates like so:

"(x1,y1),(x2,y2),(x3,y3)"

Where *x1,y1,x2,y2,x3,y3* are digits.

You should find the circle for three given points, such that the circle lies through these point and return the result as a string with the equation of the circle. In a Cartesian coordinate system (with an X and Y axis), the circle with central coordinates of (x0,y0) and radius of r can be described with the following equation:

"(x-x0)^2+(y-y0)^2=r^2"

where *x0*,*y0*,*r* are decimal numbers rounded to **two decimal points**.
Remove extraneous zeros and all decimal points, they are not necessary.
For rounding, use the standard mathematical rules.

**Input: ** Coordinates as a string..

**Output: ** The equation of the circle as a string.

**Example:**

checkio("(2,2),(6,2),(2,6)") == "(x-4)^2+(y-4)^2=2.83^2" checkio("(3,7),(6,9),(9,7)") == "(x-6)^2+(y-5.75)^2=3.25^2"

**How it is used: **
This equation, also known as Equation of the Circle,
comes from the Pythagorean theorem when applied to any point on a circle:
the radius is the hypotenuse of a right-angled triangle whose other sides are of length x − a and y − b.
Of course you can use this concept for you mathematics software, but we just want to remind about how awesome circles are.

**Precondition: **
All three given points do not lie on one line.

0 < x_{i}, y_{i}, r < 10