#include<iostream>
using namespace std;
double dp[3005][3005];
double p[3005];
int n;
int main()
{
	cin>>n;
	for(int i=1;i<=n;i++)cin>>p[i];
	for(int i=0;i<=n;i++){
		for(int j=0;j<=n/2;j++){
			if(!i)dp[i][j]=1;
			else{
				dp[i][j]=p[i]*dp[i-1][j];
				if(j>0)dp[i][j]+=(1.0-p[i])*dp[i-1][j-1];
			}
		}
	}
	printf("%.12lf",dp[n][n/2]);
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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<iostream>
using namespace std;
double dp[3005][3005];
double p[3005];
int n;
int main()
{
	cin>>n;
	for(int i=1;i<=n;i++)cin>>p[i];
	for(int i=0;i<=n;i++){
		for(int j=0;j<=n/2;j++){
			if(!i)dp[i][j]=1;
			else{
				dp[i][j]=p[i]*dp[i-1][j];
				if(j>0)dp[i][j]+=(1.0-p[i])*dp[i-1][j-1];
			}
		}
	}
	printf("%.12lf",dp[n][n/2]);
}
```