#include<bits/stdc++.h>
using namespace std;
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n;
    cin>>n;
    double head;
    vector<double> dp(n/2+1,1.0) ;
    for(int i=1;i<=n;i++){
        cin>>head;
        for(int j=n/2;j>0;j--){
            dp[j]=dp[j]*head+dp[j-1]*(1-head);
        }
        dp[0]=dp[0]*head;
    }
    cout<<setprecision(10)<<fixed<<dp[n/2]<<"\n";
}
### 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 main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n;
    cin>>n;
    double head;
    vector<double> dp(n/2+1,1.0) ;
    for(int i=1;i<=n;i++){
        cin>>head;
        for(int j=n/2;j>0;j--){
            dp[j]=dp[j]*head+dp[j-1]*(1-head);
        }
        dp[0]=dp[0]*head;
    }
    cout<<setprecision(10)<<fixed<<dp[n/2]<<"\n";
}
```