#include <bits/stdc++.h>
using namespace std;

int N;
double f[3000], p, ans;

int main() {
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin >> N;
	f[0] = 1.0;
	for(int i = 1; i <= N; ++i) {
		cin >> p;
		for(int j = i; j >= 1; --j)
			f[j] += f[j - 1] * p, f[j - 1] *= (1.0 - p);
	}
	for(int i = N / 2 + 1; i <= N; ++i)
		ans += f[i];
	cout << fixed << setprecision(10) << ans << endl;
	return 0;
}
### Prompt

Generate a cpp solution 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>
using namespace std;

int N;
double f[3000], p, ans;

int main() {
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin >> N;
	f[0] = 1.0;
	for(int i = 1; i <= N; ++i) {
		cin >> p;
		for(int j = i; j >= 1; --j)
			f[j] += f[j - 1] * p, f[j - 1] *= (1.0 - p);
	}
	for(int i = N / 2 + 1; i <= N; ++i)
		ans += f[i];
	cout << fixed << setprecision(10) << ans << endl;
	return 0;
}
```