#include<bits/stdc++.h>
using namespace std;
int main(){
  int n;
  scanf("%d",&n);
  vector<double>dp(n+1);
  dp[0]=1;
  for(int j=0 ; j<n ; j++){
    double x;
    cin>>x;
    for(int i = j+1 ; i>=0 ; i--){
      dp[i]=(i==0 ? 0 : dp[i-1])*x + dp[i]*(1-x);
    }
  }
  double sum = 0;
  for(int i=n;i>=(n+1)/2;i--){
    sum = sum + dp[i];
  }
  cout<<setprecision(10)<<sum;
  return 0;
}

### Prompt

Generate 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;
int main(){
  int n;
  scanf("%d",&n);
  vector<double>dp(n+1);
  dp[0]=1;
  for(int j=0 ; j<n ; j++){
    double x;
    cin>>x;
    for(int i = j+1 ; i>=0 ; i--){
      dp[i]=(i==0 ? 0 : dp[i-1])*x + dp[i]*(1-x);
    }
  }
  double sum = 0;
  for(int i=n;i>=(n+1)/2;i--){
    sum = sum + dp[i];
  }
  cout<<setprecision(10)<<sum;
  return 0;
}

```