Problem:
Score : 100 points

Problem Statement

Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N.
For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins.
Find the probability of having more heads than tails.

Constraints

N is an odd number.

1 \leq N \leq 2999

p_i is a real number and has two decimal places.

0 < p_i < 1

Input

Input is given from Standard Input in the following format:

N
p_1 p_2 \ldots p_N

Output

Print the probability of having more heads than tails.
The output is considered correct when the absolute error is not greater than 10^{-9}.

Sample Input 1

3
0.30 0.60 0.80

Sample Output 1

0.612

The probability of each case where we have more heads than tails is as follows:

The probability of having (Coin 1, Coin 2, Coin 3) = (Head, Head, Head) is 0.3 × 0.6 × 0.8 = 0.144;

The probability of having (Coin 1, Coin 2, Coin 3) = (Tail, Head, Head) is 0.7 × 0.6 × 0.8 = 0.336;

The probability of having (Coin 1, Coin 2, Coin 3) = (Head, Tail, Head) is 0.3 × 0.4 × 0.8 = 0.096;

The probability of having (Coin 1, Coin 2, Coin 3) = (Head, Head, Tail) is 0.3 × 0.6 × 0.2 = 0.036.

Thus, the probability of having more heads than tails is 0.144 + 0.336 + 0.096 + 0.036 = 0.612.

Sample Input 2

1
0.50

Sample Output 2

0.5

Outputs such as 0.500, 0.500000001 and 0.499999999 are also considered correct.

Sample Input 3

5
0.42 0.01 0.42 0.99 0.42

Sample Output 3

0.3821815872

Solution in Haskell (GHC 7.10.3) (judge verdict: Accepted)

{-# LANGUAGE BangPatterns #-}
{-# LANGUAGE ScopedTypeVariables #-}
import qualified Data.ByteString.Char8 as BS
import qualified Data.Vector.Unboxed as U

-- n := V.length v + 1
-- v ! i = 表が i 枚、裏が n - i 枚出る確率
solve :: [Double] -> U.Vector Double
solve ps = go ps (U.singleton 1)
  where
    go [] !vec = vec
    go (!p:ps) !vec = go ps $ U.generate (U.length vec + 1) $ \i ->
      if i == 0
      then (1 - p) * vec U.! i
      else if i == U.length vec
           then p * vec U.! (i - 1)
           else p * vec U.! (i - 1) + (1 - p) * vec U.! i

main = do
  n :: Int <- readLn
  -- 1 <= n <= 2999, n is odd
  ps :: [Double] <- map (read . BS.unpack) . BS.words <$> BS.getLine
  let result = solve ps
  -- U.length result == n + 1
  print $ U.sum $ U.drop (n `quot` 2 + 1) result