#include<bits/stdc++.h>
using namespace std;
#define MAXN 3005

double dp[MAXN];


int main(){
  int n; cin >> n;
  dp[0] = 1;
  for(int i=0; i<n; ++i) {
    double x; cin >> x;
    for(int j=i+1; j>=0; --j)
      dp[j] = ((j == 0) ? 0 : dp[j-1]*x) + (1-x)*dp[j];  
  }
  double res = 0;
  for(int i=0; i<=n; ++i){
    if(i > n-i)
      res += dp[i];
  }
  cout << fixed << setprecision(10) << res << "\n";
  return 0;
}


### Prompt

Construct a Cpp code solution to the problem outlined:

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;
#define MAXN 3005

double dp[MAXN];


int main(){
  int n; cin >> n;
  dp[0] = 1;
  for(int i=0; i<n; ++i) {
    double x; cin >> x;
    for(int j=i+1; j>=0; --j)
      dp[j] = ((j == 0) ? 0 : dp[j-1]*x) + (1-x)*dp[j];  
  }
  double res = 0;
  for(int i=0; i<=n; ++i){
    if(i > n-i)
      res += dp[i];
  }
  cout << fixed << setprecision(10) << res << "\n";
  return 0;
}


```