#include<stdio.h>
using namespace std;
double dp[3000];

int main()
{
    int N;
    scanf("%d", &N);
    double p_head;
    dp[0] = 1;
    for(int i=0; i<N; i++)
    {
        scanf("%lf", &p_head);
        for(int j = i+1; j>=0; j--)
        {
            dp[j] = (j==0?0:dp[j-1])*p_head + dp[j]*(1-p_head);
        }
    }
    double ans = 0.0;
    for(int i=N/2+1; i<=N; i++)ans += dp[i];
    printf("%.10lf\n", ans);
    return 0;
}
### Prompt

Please formulate a cpp solution 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<stdio.h>
using namespace std;
double dp[3000];

int main()
{
    int N;
    scanf("%d", &N);
    double p_head;
    dp[0] = 1;
    for(int i=0; i<N; i++)
    {
        scanf("%lf", &p_head);
        for(int j = i+1; j>=0; j--)
        {
            dp[j] = (j==0?0:dp[j-1])*p_head + dp[j]*(1-p_head);
        }
    }
    double ans = 0.0;
    for(int i=N/2+1; i<=N; i++)ans += dp[i];
    printf("%.10lf\n", ans);
    return 0;
}
```