Problem:
Score : 100 points

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

Solution in PyPy2 (5.6.0) (judge verdict: Runtime Error)

N = int(input())
P = list(map(float, input().split()))
dp = [[-1 for p in range(N)] for n in range(N+1)]
# dp[0][0] = 1
def fn(reqd, pos):
	# print("HEY")
	if pos < 0 or reqd < 0:
		return 0	
	if reqd == 0:
		dp[reqd][pos] = 1
		return 1
	if reqd > pos+1:
		dp[reqd][pos] = 0
		return 0
	else:
		if pos != 0:
			dp[reqd][pos] = round(round(round(P[pos],9)*round(fn(reqd-1,pos-1), 9), 9) + round(round(1-P[pos], 9)*round(fn(reqd,pos-1), 9), 9), 9)
			return dp[reqd][pos]
		else:
			if reqd == 1:
				dp[reqd][pos] = P[pos]
				return round(P[pos], 9)
			else:
				dp[reqd][pos] = 0
				return 0
# for reqd in range(3, 1, -1):
# 	pos = N-1
	# print("HEY")
reqd = (N+1)//2
# print(reqd)
pos = N-1
print(fn(reqd, pos))
# print(dp)
	
# print(dp)