#include <bits/stdc++.h>
using namespace std;

int main() {
    int n;
    cin>>n;
    double pi;
    vector<double> dp(n+1, 0);
    dp[0] = 1.0;
    for(int i=0; i<n; i++) {
        cin>>pi;
        for(int j = i+1; j >= 0; j--)
            dp[j] = (j == 0 ? 0 : dp[j - 1] * pi) + dp[j] * (1.0 - pi);
    }
    cout<< fixed << setprecision(10) <<
        accumulate(dp.begin() + (n+1)/2, dp.end(), 0.0) <<endl;
    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 <bits/stdc++.h>
using namespace std;

int main() {
    int n;
    cin>>n;
    double pi;
    vector<double> dp(n+1, 0);
    dp[0] = 1.0;
    for(int i=0; i<n; i++) {
        cin>>pi;
        for(int j = i+1; j >= 0; j--)
            dp[j] = (j == 0 ? 0 : dp[j - 1] * pi) + dp[j] * (1.0 - pi);
    }
    cout<< fixed << setprecision(10) <<
        accumulate(dp.begin() + (n+1)/2, dp.end(), 0.0) <<endl;
    return 0;
}

```