#include<bits/stdc++.h>
using namespace std;
double dp[3200];
int main()
{
	int t;
	scanf("%d",&t);
	dp[0]=1;
	for(int i = 0;i < t;i++)
	{
		double p,ps;
		scanf("%lf",&p);
		ps=1;
		ps-=p;
		for(int j = i+1;j>=0;j--)
		{
			double pp=dp[j];
			dp[j+1]+=pp*p;
			dp[j]=0;
			dp[j]=pp*(ps);
		}
	}
	double cc=0;
	for(int i = 0;i <= t;i++)
	{
		if(i>t/2)
		cc+=dp[i];
	}
	printf("%.12f",cc);
}
### Prompt

Create a solution in cpp for 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;
double dp[3200];
int main()
{
	int t;
	scanf("%d",&t);
	dp[0]=1;
	for(int i = 0;i < t;i++)
	{
		double p,ps;
		scanf("%lf",&p);
		ps=1;
		ps-=p;
		for(int j = i+1;j>=0;j--)
		{
			double pp=dp[j];
			dp[j+1]+=pp*p;
			dp[j]=0;
			dp[j]=pp*(ps);
		}
	}
	double cc=0;
	for(int i = 0;i <= t;i++)
	{
		if(i>t/2)
		cc+=dp[i];
	}
	printf("%.12f",cc);
}
```