#include<bits/stdc++.h>
using namespace std;
#define maxn 3000
int main()
{
	int n;
	scanf("%d",&n);
	double dp[maxn] = {1.0};
	for(int i=1; i<=n; i++)
	{
		double p;
		scanf("%lf",&p);
		for(int j=min(i,n/2); j>0; j--)
			dp[j] = dp[j]*(1-p) + dp[j-1]*p;
		dp[0] = dp[0]*(1-p);
	}
	double ans = 1;
	for(int i=0; i<=n/2; i++)
		ans -= dp[i];
	printf("%.9f\n",ans);
	return 0;
}
### 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;
#define maxn 3000
int main()
{
	int n;
	scanf("%d",&n);
	double dp[maxn] = {1.0};
	for(int i=1; i<=n; i++)
	{
		double p;
		scanf("%lf",&p);
		for(int j=min(i,n/2); j>0; j--)
			dp[j] = dp[j]*(1-p) + dp[j-1]*p;
		dp[0] = dp[0]*(1-p);
	}
	double ans = 1;
	for(int i=0; i<=n/2; i++)
		ans -= dp[i];
	printf("%.9f\n",ans);
	return 0;
}
```