/*
 *  Lang    : C++
 *  Author  : little_angel
 */
#include <bits/stdc++.h>

using namespace std;
double dp[100010], sum;

signed main() {
    int n;
    cin >> n;
    dp[0] = 1;
    for (int i = 1; i <= n; ++i) {
        double p;
        cin >> p;
        for (int j = i; j >= 0; --j) {
            dp[j] = (j ? dp[j - 1] : 0) * p + dp[j] * (1 - p);
        }
    }
    for (int i = n; n - i < i; --i) sum += dp[i];
    printf("%0.10lf", sum);
}
### 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
/*
 *  Lang    : C++
 *  Author  : little_angel
 */
#include <bits/stdc++.h>

using namespace std;
double dp[100010], sum;

signed main() {
    int n;
    cin >> n;
    dp[0] = 1;
    for (int i = 1; i <= n; ++i) {
        double p;
        cin >> p;
        for (int j = i; j >= 0; --j) {
            dp[j] = (j ? dp[j - 1] : 0) * p + dp[j] * (1 - p);
        }
    }
    for (int i = n; n - i < i; --i) sum += dp[i];
    printf("%0.10lf", sum);
}
```