#include <bits/stdc++.h>
using namespace std;
using ld = long double;

ld dp[3002];
int n;

int main() {
	cin>>n;
	dp[0] = 1;
	for(int i = 0; i < n; ++i){
        ld x; cin>>x;
        for(int j = i+1; j >= 0; --j){
            dp[j] = (j == 0 ? 0 : dp[j-1] * x) + (1 - x) * dp[j]; 
        }
	}
	ld ans = 0;
	for(int i = n/2+1; i <= n; i++){
	    ans += dp[i];
	}
	cout.precision(10);
	cout<<ans<<fixed;
	return 0;
}
### Prompt

Develop a solution in CPP 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;
using ld = long double;

ld dp[3002];
int n;

int main() {
	cin>>n;
	dp[0] = 1;
	for(int i = 0; i < n; ++i){
        ld x; cin>>x;
        for(int j = i+1; j >= 0; --j){
            dp[j] = (j == 0 ? 0 : dp[j-1] * x) + (1 - x) * dp[j]; 
        }
	}
	ld ans = 0;
	for(int i = n/2+1; i <= n; i++){
	    ans += dp[i];
	}
	cout.precision(10);
	cout<<ans<<fixed;
	return 0;
}
```