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