Let $N$ be a positive odd number.

There are $N$ coins, numbered $1,2,\dots ,N$. For each $i$ $(1\le i\le N)$, 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.
• $1\le N\le 2999$.
• $p_i$ is a real number and has two decimal places.
• $0<p_i<1$

Input Specification

The first line will contain the integer $N$.

The next line will contain $N$ floats, $p_1,p_2,\dots ,p_N$.

Output Specification

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

Explanation For Sample 1

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

• The probability of having $(Coin1,Coin2,Coin3)=(Head,Head,Head)$ is $0.3\times 0.6\times 0.8=0.144$;
• The probability of having $(Coin1,Coin2,Coin3)=(Tail,Head,Head)$ is $0.7\times 0.6\times 0.8=0.336$;
• The probability of having $(Coin1,Coin2,Coin3)=(Head,Tail,Head)$ is $0.3\times 0.4\times 0.8=0.096$;
• The probability of having $(Coin1,Coin2,Coin3)=(Head,Head,Tail)$ is $0.3\times 0.6\times 0.2=0.036$;

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

Explanation For Sample 2

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