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