LED Series Resistor

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

An LED with forward voltage $V_F$ needs a series resistor $R = \frac{V_S - V_F}{I}$ to limit its current. Print the exact $R$ (1 decimal) and the smallest E12 standard value that is at least $R$ (so the current is never above the limit).

E12 values are $1.0, 1.2, 1.5, 1.8, 2.2, 2.7, 3.3, 3.9, 4.7, 5.6, 6.8, 8.2$ times $10^0 \dots 10^6$ .

Input

$V_S$ (volts), $V_F$ (volts) and $I$ (milliamps).

Output

Exact resistance and the chosen E12 value (printed without trailing zeros, e.g. 330 or 1500).

Constraints

$V_F < V_S \le 300$ , $1 \le I \le 1000$ mA

Sample 1

Input

5 2 10

Output

300.0 330

Sample 2

Input

12 3.2 20

Output

440.0 470