Resistor Power Rating

Easy#engineering #electronics· solved by 0 · time limit 2.0 s

A resistor $R$ with voltage $V$ across it dissipates $P = \frac{V^2}{R}$ watts. Engineers pick a rating at least twice $P$ . Print $P$ (3 decimals) and the smallest standard rating from 0.125, 0.25, 0.5, 1, 2, 5, 10 W that is $\ge 2P$ (as 0.25W, 1W, …), or Too much if none is enough.

Input

$V$ and $R$ .

Output

$P$ and the rating.

Constraints

$0 < V \le 1000$ , $1 \le R \le 10^7$

Sample 1

Input

5 220

Output

0.114 0.25W

Sample 2

Input

12 10

Output

14.400 Too much