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

int main(void){
	int n;
	cin>>n;
	double *DP=new double[n+1];
	DP[0]=1;
	for(int i=1;i<=n;i++) DP[i]=0;

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

	double ans=0;
	for(int i=n;i>n/2;i--){
		ans+=DP[i];
	}

	printf("%.10f\n",ans);
	
	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 main(void){
	int n;
	cin>>n;
	double *DP=new double[n+1];
	DP[0]=1;
	for(int i=1;i<=n;i++) DP[i]=0;

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

	double ans=0;
	for(int i=n;i>n/2;i--){
		ans+=DP[i];
	}

	printf("%.10f\n",ans);
	
	return 0;
}

```