#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
int n;
double p, dp[2][3004];

int main() {
	cin>>n;
	dp[0][0] = 1.0;
	for(int i = 1;i<=n;i++){
		cin>>p;
		for(int j = 0;j<=i;j++)
			dp[i%2][j] = dp[(i-1)%2][j]*(1-p)+dp[(i-1)%2][j-1]*p;
	}
	double ans = 0;
	for(int i = (n+1)/2;i<=n;i++)
		ans += dp[n%2][i];
	printf("%.10f",ans);
	return 0;
}
### Prompt

Your task is to create 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;
const int mod = 1000000007;
int n;
double p, dp[2][3004];

int main() {
	cin>>n;
	dp[0][0] = 1.0;
	for(int i = 1;i<=n;i++){
		cin>>p;
		for(int j = 0;j<=i;j++)
			dp[i%2][j] = dp[(i-1)%2][j]*(1-p)+dp[(i-1)%2][j-1]*p;
	}
	double ans = 0;
	for(int i = (n+1)/2;i<=n;i++)
		ans += dp[n%2][i];
	printf("%.10f",ans);
	return 0;
}
```