#include <bits/stdc++.h>

using namespace std;

int main() {
  int n;
  cin >> n;
  vector<double> dp(n+1);
  double p;

  dp[0] = 1; //probability of i heads so far
  for (int i=0; i<n; ++i) {
    cin >> p;
    for (int j=i+1; j>=0; --j) {
      dp[j] = (!j ? 0 : dp[j-1] * p) + dp[j] * (1-p);
    }
  }

  double ans = 0;
  for (int i=n; i>n/2; --i)
    ans += dp[i];

  cout.precision(10);
  cout << ans << '\n';

  return 0;
}

### Prompt

Please provide a Cpp coded solution 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 <bits/stdc++.h>

using namespace std;

int main() {
  int n;
  cin >> n;
  vector<double> dp(n+1);
  double p;

  dp[0] = 1; //probability of i heads so far
  for (int i=0; i<n; ++i) {
    cin >> p;
    for (int j=i+1; j>=0; --j) {
      dp[j] = (!j ? 0 : dp[j-1] * p) + dp[j] * (1-p);
    }
  }

  double ans = 0;
  for (int i=n; i>n/2; --i)
    ans += dp[i];

  cout.precision(10);
  cout << ans << '\n';

  return 0;
}

```