Proficiency

Yeongseon has finished shopping and walks to the checkout. There are two counters. $L_1$ people stand in line at counter 1 and $L_2$ people stand in line at counter 2. Each clerk serves the person at the front of the line, one person at a time.

The time a clerk needs for one person depends on the clerk's proficiency. A clerk with proficiency $p$ finishes one person in exactly $k$ seconds with probability $\frac{1}{p}\left(1 - \frac{1}{p}\right)^{k-1}$. The clerk at counter 1 has proficiency $P_1$, the clerk at counter 2 has proficiency $P_2$, and the times are independent across people.

Given $L_1$, $L_2$, $P_1$, $P_2$, compute the probability that standing in line at counter 1 is better than standing in line at counter 2. That is the probability that the last person in line 1 finishes before the last person in line 2. Finishing in the same second does not count as finishing first.

The first line contains $L_1$, $L_2$, $P_1$, $P_2$ separated by spaces. ($1 \le L_1, L_2, P_1, P_2 \le 1000$)

Print the probability that the last person at counter 1 finishes before the last person at counter 2, rounded to nine decimal places. Always print exactly nine digits after the decimal point. If $0^0$ comes up while computing the probability, treat it as 1.