#include <bits/stdc++.h>

using namespace std;

int n;
double p[3100];
double dp[3100][3100];
int main()
{
    cin >> n;
    for(int i = 0 ; i < n ; i++)cin >> p[i];

    for(int i = n/2 + 1 ; i <= n; i++)dp[n][i] = 1;
    for(int i = n-1 ; i >= 0 ; i--)
        for(int cnt = 0 ; cnt <= i ;cnt ++ )
            dp[i][cnt] = p[i]*dp[i+1][cnt+1] + (1-p[i])*dp[i+1][cnt];

    printf("%.10f\n",dp[0][0]);
    return 0;
}

### Prompt

Create a solution in CPP for 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 n;
double p[3100];
double dp[3100][3100];
int main()
{
    cin >> n;
    for(int i = 0 ; i < n ; i++)cin >> p[i];

    for(int i = n/2 + 1 ; i <= n; i++)dp[n][i] = 1;
    for(int i = n-1 ; i >= 0 ; i--)
        for(int cnt = 0 ; cnt <= i ;cnt ++ )
            dp[i][cnt] = p[i]*dp[i+1][cnt+1] + (1-p[i])*dp[i+1][cnt];

    printf("%.10f\n",dp[0][0]);
    return 0;
}

```