#include <bits/stdc++.h>

int main()
{
	int n;
	scanf("%d", &n);
	std::vector<long double> prob(n + 1);
	prob[0] = 1.0;
	for (int i{}; i < n; i++)
	{
		long double p;
		std::cin >> p;
		for (int j{i + 1}; j > 0; j--)
			prob[j] = p * prob[j - 1] + (1.0 - p) * prob[j];
		prob[0] *= 1.0 - p;
	}
	long double ans{};
	for (int i{n / 2 + 1}; i <= n; i++)
		ans += prob[i];
	printf("%.10Lf\n", ans);

	return 0;
}
### Prompt

Please create a solution in cpp to the following problem:

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}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

### Response

```cpp
#include <bits/stdc++.h>

int main()
{
	int n;
	scanf("%d", &n);
	std::vector<long double> prob(n + 1);
	prob[0] = 1.0;
	for (int i{}; i < n; i++)
	{
		long double p;
		std::cin >> p;
		for (int j{i + 1}; j > 0; j--)
			prob[j] = p * prob[j - 1] + (1.0 - p) * prob[j];
		prob[0] *= 1.0 - p;
	}
	long double ans{};
	for (int i{n / 2 + 1}; i <= n; i++)
		ans += prob[i];
	printf("%.10Lf\n", ans);

	return 0;
}
```