#include <bits/stdc++.h>
#define N 3001
using namespace std;

int n;
double p, ans, dp[N];

int main() {
	cin>>n;
	dp[0]=1;
	for (int i=1; i<=n; i++) {
		cin>>p;
		for (int j=n; j>0; j--) dp[j]=dp[j-1]*p+dp[j]*(1-p);
		dp[0]*=1-p;
	}
	for (int i=n; i>n/2; i--) ans+=dp[i];
	printf("%.10f\n",ans);
	return 0;
}
### Prompt

In cpp, your task is to solve 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>
#define N 3001
using namespace std;

int n;
double p, ans, dp[N];

int main() {
	cin>>n;
	dp[0]=1;
	for (int i=1; i<=n; i++) {
		cin>>p;
		for (int j=n; j>0; j--) dp[j]=dp[j-1]*p+dp[j]*(1-p);
		dp[0]*=1-p;
	}
	for (int i=n; i>n/2; i--) ans+=dp[i];
	printf("%.10f\n",ans);
	return 0;
}
```