#include <bits/stdc++.h>

using namespace std;

double dp[3005][3005];

int main(){
    int N;
    cin >> N;
    dp[0][0] = 1;
    for(int i = 1; i<=N; i++){
        double p;
        cin >> p;
        for(int j = 0; j<=N; j++){
            dp[i][j] += dp[i-1][j]*(1-p);
            dp[i][j+1] += dp[i-1][j]*p;
        }
    }
    double tot = 0;
    for(int i = (N+1)/2; i<=N; i++){
        tot += dp[N][i];
    }
    printf("%.9f", tot);
}

### 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;

double dp[3005][3005];

int main(){
    int N;
    cin >> N;
    dp[0][0] = 1;
    for(int i = 1; i<=N; i++){
        double p;
        cin >> p;
        for(int j = 0; j<=N; j++){
            dp[i][j] += dp[i-1][j]*(1-p);
            dp[i][j+1] += dp[i-1][j]*p;
        }
    }
    double tot = 0;
    for(int i = (N+1)/2; i<=N; i++){
        tot += dp[N][i];
    }
    printf("%.9f", tot);
}

```