#include<cstdio>
double dp[3000][3000];
int main(){
  int n;
  scanf("%d",&n);
  dp[0][0]=1;
  for(int i=1;i<=n;i++){
    double p;
    scanf("%lf",&p);
    dp[i][0]=dp[i-1][0]*(1-p);
    for(int j=1;j<i;j++)dp[i][j]=dp[i-1][j-1]*p+dp[i-1][j]*(1-p);
    dp[i][i]=dp[i-1][i-1]*p;
  }
  double ans=0;
  for(int i=n;i>n/2;i--)ans+=dp[n][i];
  printf("%.10lf",ans);
  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<cstdio>
double dp[3000][3000];
int main(){
  int n;
  scanf("%d",&n);
  dp[0][0]=1;
  for(int i=1;i<=n;i++){
    double p;
    scanf("%lf",&p);
    dp[i][0]=dp[i-1][0]*(1-p);
    for(int j=1;j<i;j++)dp[i][j]=dp[i-1][j-1]*p+dp[i-1][j]*(1-p);
    dp[i][i]=dp[i-1][i-1]*p;
  }
  double ans=0;
  for(int i=n;i>n/2;i--)ans+=dp[n][i];
  printf("%.10lf",ans);
  return 0;
}
```