Go to school

Tongjiang wanna go back to UESTC from Xipu. He had arrived Xipu bus station at $(T=0)$.He would like take a bus or walk, for details, it takes $A$ minutes to take a bus while $B$ minutes to walk($A<B$ will be hold).There are $N$ bus lines all can take Tongjiang to UESTC, But as it was getting late,some bus lines may had already stopped running. Properly speaking, the bus line have a probability of $(1-\frac{L_i}{M})$ that it had already stopped running and no more bus will come to Xipu station, and a probability of $(\frac{L_i}{M}$) that there will be a bus come to Xipu station to take Tongjiang to UESTC. And in the second case the bus may arrive at bus station at any time $T$, which holds $(0 \leq T \leq Li)$, with equal probability. Tongjiang is a smart boy, he had $Q$ different strategy to wait bus. Under strategy $i$ He would set a latest time point $T_i$ first. Then he started waiting for bus and take the first bus come to station who can take him to school or when time come to $(T=T_i)$ and still no bus come, he would start walk to school.$\\$
Now, Tongjiang wonders under each strategy, what's the expect time he arrive at school.

For all bus line $i$, $( \frac{1}{3} M \leq L_i \leq \frac{2}{3} M )$

For all strategy $i$, $( Min(L_1,L_2...L_N) \leq T_i \leq Max(L_1,L_2...L_N) )$

$(6 \leq M \leq 100000)$

$(1 \leq A,B\leq 100000)$

$(2 \leq N \leq 12)$

$(1 \leq Q \leq 100000 )$

First line five positive integer $N,Q,M,A$ and $B$.
Second line N positive integer $(L_1,L_2...L_N)$
Then $Q$ lines following,each line contains one integer means a latest time point $T_i$.

For each strategy,print one line with you answer.
Your answer will be considered to be right if
$\min{\left\{\left|yourans-stdans\right|, \left|\frac{yourans-stdans}{stdans}\right|\right\}} < 10^{-4}$

$(min(abs(yourans-stdans),abs(\frac{yourans-stdans}{stdans})<1e-4)$.

2 2 6 1 2
2 4
2
3

2.851852
3.129630

2018 Multi-University Training Contest 7