#include <iostream>
using namespace std;

int n;
double p[3300], dp[3300];

int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> p[i];

  dp[0] = 1;
  for (double prob : p) {
    for (int j = n; 0 <= j; j--) {
      dp[j] = dp[j] * (1 - prob) + dp[j - 1] * prob;
    }
  }
  double ans = 0;
  for (int i = (n + 1) / 2; i <= n; i++) {
    ans += dp[i];
  }

  printf("%.10f\n", ans);
  return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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 <iostream>
using namespace std;

int n;
double p[3300], dp[3300];

int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> p[i];

  dp[0] = 1;
  for (double prob : p) {
    for (int j = n; 0 <= j; j--) {
      dp[j] = dp[j] * (1 - prob) + dp[j - 1] * prob;
    }
  }
  double ans = 0;
  for (int i = (n + 1) / 2; i <= n; i++) {
    ans += dp[i];
  }

  printf("%.10f\n", ans);
  return 0;
}
```