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

int main(){
    cout.precision(10);
    int n;
    cin>>n;
    vector<double> dp(n+1);
    cin>>dp[0];
    dp[1] = 1-dp[0];
    for(int i=1; i<n; ++i){
        double h,t;
        cin>>h;
        t=1-h;
        for(int j=i; j>=0; --j){
            dp[j+1]+=dp[j]*t;
            dp[j]=dp[j]*h;
        }
    }
    
    double ans=0;
    for(int i=0; i<=(n-1)/2; ++i) ans+=dp[i];
    cout<<ans<<"\n";
}
### 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;

int main(){
    cout.precision(10);
    int n;
    cin>>n;
    vector<double> dp(n+1);
    cin>>dp[0];
    dp[1] = 1-dp[0];
    for(int i=1; i<n; ++i){
        double h,t;
        cin>>h;
        t=1-h;
        for(int j=i; j>=0; --j){
            dp[j+1]+=dp[j]*t;
            dp[j]=dp[j]*h;
        }
    }
    
    double ans=0;
    for(int i=0; i<=(n-1)/2; ++i) ans+=dp[i];
    cout<<ans<<"\n";
}
```