#include<bits/stdc++.h>

using namespace std;

const int N = 3e3 + 3;
double dp[N];

int main()
{
	int n;
	cin >> n;

	dp[0] = 1;
	double p;
	for(int i = 1; i <= n; i++){
		cin >> p;
		for(int j = i; j >= 0; j--)
			dp[j] = (j == 0 ? 0 : dp[j - 1] * p) + dp[j] * (1 - p);
	}

	double ans = 0;
	for(int i = 0; i <= n; i++)
		if(i > n - i)
			ans += dp[i];

	cout << fixed << setprecision(10);	
	cout << ans << "\n";
}
### 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>

using namespace std;

const int N = 3e3 + 3;
double dp[N];

int main()
{
	int n;
	cin >> n;

	dp[0] = 1;
	double p;
	for(int i = 1; i <= n; i++){
		cin >> p;
		for(int j = i; j >= 0; j--)
			dp[j] = (j == 0 ? 0 : dp[j - 1] * p) + dp[j] * (1 - p);
	}

	double ans = 0;
	for(int i = 0; i <= n; i++)
		if(i > n - i)
			ans += dp[i];

	cout << fixed << setprecision(10);	
	cout << ans << "\n";
}
```