#include <bits/stdc++.h>
using namespace std;
#define ll long long
const ll INF= 1e18L+5;

int main() {
	int n;
	cin >> n;
	std::vector<double> dp(n+1,0);
	dp[0]=1;
	for(int i=0; i<n; i++){
		double x;
		cin>> x;
		for(int j=i+1;j>=0; j--){
			dp[j] = (j==0 ? 0: dp[j-1]*x)+dp[j]*(1-x);
		}
	}
	double ans = 0;
	for(int h =n/2+1; h<=n;h++)
		ans += dp[h];
	printf("%0.10f\n",ans);
}
### 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 <bits/stdc++.h>
using namespace std;
#define ll long long
const ll INF= 1e18L+5;

int main() {
	int n;
	cin >> n;
	std::vector<double> dp(n+1,0);
	dp[0]=1;
	for(int i=0; i<n; i++){
		double x;
		cin>> x;
		for(int j=i+1;j>=0; j--){
			dp[j] = (j==0 ? 0: dp[j-1]*x)+dp[j]*(1-x);
		}
	}
	double ans = 0;
	for(int h =n/2+1; h<=n;h++)
		ans += dp[h];
	printf("%0.10f\n",ans);
}
```