A stone is dropped into a well. $t$ seconds later, a splash is heard. The acceleration due to gravity is $g$ metres per second per second, and the speed of sound is $c$ metres per second. Calculate the depth of the well and the downwards velocity with which the stone hits the water.

Input Specification

Line $1$: The value of $g$ $(0<g\le 1000)$
Line $2$: The value of $c$ $(0<c\le 1000)$
Line $3$: The value of $t$ $(0<t\le 1000)$

Output Specification

Line $1$: The depth of the well in metres.
Line $2$: The speed of the stone as it hits the water, in metres per second.

Sample Input

9.80
334
3.41

Sample Output

51.90
31.90

Your answers must have a relative or absolute error of less than $0.01$.

Notes

The following equations may be useful:

$${\displaystyle \begin{array}{rl}\mathrm{\Delta }\overrightarrow{d}& =\overrightarrow{v}\mathrm{\Delta }t\\ {\overrightarrow{v}}_2-{\overrightarrow{v}}_1& =\overrightarrow{a}\mathrm{\Delta }t\\ \mathrm{\Delta }\overrightarrow{d}& =\frac{1}{2}({\overrightarrow{v}}_1+{\overrightarrow{v}}_2)\mathrm{\Delta }t\\ \mathrm{\Delta }\overrightarrow{d}& ={\overrightarrow{v}}_1\mathrm{\Delta }t+\frac{1}{2}\overrightarrow{a}(\mathrm{\Delta }t{)}^2\\ \mathrm{\Delta }\overrightarrow{d}& ={\overrightarrow{v}}_2\mathrm{\Delta }t-\frac{1}{2}\overrightarrow{a}(\mathrm{\Delta }t{)}^2\\ \Vert {\overrightarrow{v}}_2{\Vert }^2-\Vert {\overrightarrow{v}}_1{\Vert }^2& =2\overrightarrow{a}\cdot \mathrm{\Delta }\overrightarrow{d}\end{array}}$$

Here $\mathrm{\Delta }t,\mathrm{\Delta }\overrightarrow{d},\overrightarrow{v},{\overrightarrow{v}}_1,{\overrightarrow{v}}_2,\overrightarrow{a}$ represent time, displacement, velocity, initial velocity, final velocity, and acceleration, respectively.