These problems are from the atcoder DP contest, and were transferred onto DMOJ. All problem statements were made by several atcoder users. As there is no access to the test data, all data is randomly generated. If there are issues with the statement or data, please contact Rimuru or Ninjaclasher on slack.

Let ~N~ be a positive odd number.

There are ~N~ coins, numbered ~1, 2, \ldots, 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, \ldots, 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