#include<bits/stdc++.h>
using namespace std;
using LL=long long;
int main(){
  int N;cin>>N;
  vector<double>dp(N+1,0);
  dp[0]=1.0;
  for(int i=1;i<=N;i++){
    double p;cin>>p;
    for(int j=i;j>=1;j--)
      dp[j]=dp[j]*p+dp[j-1]*(1.0-p);
    dp[0]=dp[0]*p;
  }
  cout<<setprecision(10)<<accumulate(dp.begin(),dp.begin()+(N+1)/2,0.0)<<endl;
}
### Prompt

Your task is to create a Cpp solution to the following problem:

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 LL=long long;
int main(){
  int N;cin>>N;
  vector<double>dp(N+1,0);
  dp[0]=1.0;
  for(int i=1;i<=N;i++){
    double p;cin>>p;
    for(int j=i;j>=1;j--)
      dp[j]=dp[j]*p+dp[j-1]*(1.0-p);
    dp[0]=dp[0]*p;
  }
  cout<<setprecision(10)<<accumulate(dp.begin(),dp.begin()+(N+1)/2,0.0)<<endl;
}
```