output_description: Print the probability of having more heads than tails. The output is
  considered correct when the absolute error is not greater than 10^{-9}.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
      p_1 p_2 \ldots p_N
attempt: from subprocess import *
  
  call(
      (
          "pypy3",
          "-c",
          """
  n=int(input())
  p=list(map(float,input().split()))
  dp=[[0]*-~n for _ in range(n)]
  dp[0][1]=p[0]
  dp[0][0]=1-p[0]
  for i in range(1,n):
    dp[i][0]=dp[i-1][0]*(1-p[i])
    for j in range(1,n+1):
      dp[i][j]=dp[i-1][j-1]*p[i]+dp[i-1][j]*(1-p[i])
  print(sum(dp[-1][-~n//2:]))
  """,
      )
  )
  
problem_description: 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.
samples: [{"input": "3\n    0.30 0.60 0.80", "output": "0.612\n    \n\nThe probability of each case where we have more heads than tails is as\nfollows:\n\n  * The probability of having (Coin 1, Coin 2, Coin 3) = (Head, Head, Head) is 0.3 \u00d7 0.6 \u00d7 0.8 = 0.144;\n  * The probability of having (Coin 1, Coin 2, Coin 3) = (Tail, Head, Head) is 0.7 \u00d7 0.6 \u00d7 0.8 = 0.336;\n  * The probability of having (Coin 1, Coin 2, Coin 3) = (Head, Tail, Head) is 0.3 \u00d7 0.4 \u00d7 0.8 = 0.096;\n  * The probability of having (Coin 1, Coin 2, Coin 3) = (Head, Head, Tail) is 0.3 \u00d7 0.6 \u00d7 0.2 = 0.036.\n\nThus, the probability of having more heads than tails is 0.144 + 0.336 + 0.096\n+ 0.036 = 0.612.\n\n* * *"}, {"input": "1\n    0.50", "output": "0.5\n    \n\nOutputs such as `0.500`, `0.500000001` and `0.499999999` are also considered\ncorrect.\n\n* * *"}, {"input": "5\n    0.42 0.01 0.42 0.99 0.42", "output": "0.3821815872"}]