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

int n;
long double prob[3002];

int main(){
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	prob[0] = 1;
	cin>>n;
	for(int i=1;i<=n;i++){
		long double p;
		cin>>p;
		for(int j=n;j>0;j--){
			prob[j] = prob[j]*(1-p) + prob[j-1]*p;
		}
		prob[0] *= (1-p);
	}
	long double ans = 0.0;
	for(int i=n/2+1;i<=n;i++){
		ans += prob[i];
	}
	cout<<setprecision(12);
	cout<<ans;
	return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;

int n;
long double prob[3002];

int main(){
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	prob[0] = 1;
	cin>>n;
	for(int i=1;i<=n;i++){
		long double p;
		cin>>p;
		for(int j=n;j>0;j--){
			prob[j] = prob[j]*(1-p) + prob[j-1]*p;
		}
		prob[0] *= (1-p);
	}
	long double ans = 0.0;
	for(int i=n/2+1;i<=n;i++){
		ans += prob[i];
	}
	cout<<setprecision(12);
	cout<<ans;
	return 0;
}

```