#include <cstdio>
#include <numeric>

double dp[3000];

int main()
{
  int n;
  scanf("%d", &n);
  double p[2999];
  for (int i = 0; i < n; ++i)
  {
    scanf("%lf", p + i);
  }
  dp[0] = 1.;
  for (int i = 0; i < n; ++i)
  {
    for (int j = i + 1; j >= 0; --j)
    {
      dp[j] = dp[j] * (1 - p[i]) + (j > 0 ? dp[j - 1] * p[i] : 0);
    }
  }
  printf("%.16f\n", std::accumulate(dp + n / 2 + 1, dp + n + 1, 0.));
  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 <cstdio>
#include <numeric>

double dp[3000];

int main()
{
  int n;
  scanf("%d", &n);
  double p[2999];
  for (int i = 0; i < n; ++i)
  {
    scanf("%lf", p + i);
  }
  dp[0] = 1.;
  for (int i = 0; i < n; ++i)
  {
    for (int j = i + 1; j >= 0; --j)
    {
      dp[j] = dp[j] * (1 - p[i]) + (j > 0 ? dp[j - 1] * p[i] : 0);
    }
  }
  printf("%.16f\n", std::accumulate(dp + n / 2 + 1, dp + n + 1, 0.));
  return 0;
}
```