#include "bits/stdc++.h"
using namespace std;
int N; double p, dp[3000][3000], ans;
int main(){
    scanf("%d", &N); dp[0][0] = 1.0;
    for(int i=1; i<=N; i++){
        scanf("%lf", &p);
        dp[i][0] = dp[i-1][0] * (1-p);
        for(int j=1; j<=i; j++){
            dp[i][j] = dp[i-1][j-1]*p + dp[i-1][j]*(1-p);
        }
    }
    for(int i=N; i>N/2; i--) ans += dp[N][i];
    printf("%.9f\n", 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 N; double p, dp[3000][3000], ans;
int main(){
    scanf("%d", &N); dp[0][0] = 1.0;
    for(int i=1; i<=N; i++){
        scanf("%lf", &p);
        dp[i][0] = dp[i-1][0] * (1-p);
        for(int j=1; j<=i; j++){
            dp[i][j] = dp[i-1][j-1]*p + dp[i-1][j]*(1-p);
        }
    }
    for(int i=N; i>N/2; i--) ans += dp[N][i];
    printf("%.9f\n", ans);
}
```