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      #-}

import Prelude as P
import Data.List
import Data.Vector.Unboxed as V hiding(sum,forM,forM_,map)
import Data.Array.Unboxed  as A
import Data.Array.MArray
import Data.Array.ST
import Control.Monad
import Control.Monad.ST

type P = Vector Double
type DP s = STUArray s (Int,Int) Double

setArr :: DP s -> P -> Int -> ST s ()
setArr dparr ps n = do
  let p i  = ps V.! (i-1)
      p_ i = 1-(p i)
  forM_ [1..n] $ \i -> writeArray dparr (0,i) $ (V.scanl (\x y -> x*(1-y)) 1 ps) V.! i
  forM_ [1..div (n-1) 2] $ \i -> do
    writeArray dparr (i,i)   $ (V.scanl (\x y -> x*y) 1 ps) V.! i

dp_i :: DP s -> P -> Int -> Int -> ST s ()
dp_i dparr ps k i = do
  let p i  = ps V.! (i-1)
      p_ i = 1-(p i)
  ek  <- readArray dparr (k,i-1)
  ek1 <- readArray dparr (k-1,i-1)
  writeArray dparr (k,i) $ ek*(p_ i) + ek1*(p i)

solve :: P -> Int -> ST s Double
solve ps n = do
  dparr <- newArray ((0,1),(div (n-1) 2,n)) 0
  setArr dparr ps n
  forM_ [1..div (n-1) 2] $ \k ->
   forM_ [k+1..n] $ \i -> dp_i dparr ps k i
  ans0 <- forM [0..div (n-1) 2] $ \k -> readArray dparr (k,n)
  return $ 1-(sum ans0)

main :: IO ()
main = do
  n <- read <$> getLine
  ps <- map read.words <$> getLine
  putStrLn $ show $ runST $ solve (V.fromList ps)  n

mod0 :: Int -> Int
mod0 n = mod n 1000000007