#include<bits/stdc++.h>

double dp[3000][3000];

int main()
{
  int n;
  double i,ans=0;

  scanf("%d",&n);
  dp[0][0]=1;
  for(int c=1;c<=n;c++)
  {
    scanf("%lf",&i);
    for(int c2=0;c2<=c;c2++)
    dp[c][c2]=(c2==0?0:(i*dp[c-1][c2-1]))+((1-i)*dp[c-1][c2]);
  }
  for(int c=(n+1)/2;c<=n;c++)ans+=dp[n][c];
  printf("%.10lf\n",ans);
}

### Prompt

Please formulate 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>

double dp[3000][3000];

int main()
{
  int n;
  double i,ans=0;

  scanf("%d",&n);
  dp[0][0]=1;
  for(int c=1;c<=n;c++)
  {
    scanf("%lf",&i);
    for(int c2=0;c2<=c;c2++)
    dp[c][c2]=(c2==0?0:(i*dp[c-1][c2-1]))+((1-i)*dp[c-1][c2]);
  }
  for(int c=(n+1)/2;c<=n;c++)ans+=dp[n][c];
  printf("%.10lf\n",ans);
}

```