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

int main(){
    int n; cin>>n;
    double p[n];
    for(int i=1;i<=n;i++) cin>>p[i];
    double dp[n+1][n+1];
    dp[0][0] = 1;
    for(int i=1;i<=n;i++){
        for(int j=0;j<=i;j++){
            dp[i][j] = dp[i-1][j]*(1-p[i]) + dp[i-1][j-1]*p[i];
        }
    }
    double ans=0.0;
    for(int j=n/2+1;j<=n;j++){
        ans += dp[n][j];
    }

    cout << fixed << setprecision(10) << ans;
}
### Prompt

Please create a solution in CPP 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>
using namespace std;

int main(){
    int n; cin>>n;
    double p[n];
    for(int i=1;i<=n;i++) cin>>p[i];
    double dp[n+1][n+1];
    dp[0][0] = 1;
    for(int i=1;i<=n;i++){
        for(int j=0;j<=i;j++){
            dp[i][j] = dp[i-1][j]*(1-p[i]) + dp[i-1][j-1]*p[i];
        }
    }
    double ans=0.0;
    for(int j=n/2+1;j<=n;j++){
        ans += dp[n][j];
    }

    cout << fixed << setprecision(10) << ans;
}
```