#include <bits/stdc++.h>
using namespace std;

int n;
double ans;

int main() {
	cin >> n;
	double p[n + 1], dp[n + 1][n + 1];
	dp[0][0] = 1;
	for (int i=1; i<=n; i++) {
		cin >> p[i];
		dp[i][0] = dp[i-1][0] * (1 - p[i]);
		for (int j=1; j<i; j++) {
			dp[i][j] = dp[i-1][j-1] * p[i] + dp[i-1][j] * (1 - p[i]);
		}
		dp[i][i] = dp[i-1][i-1] * p[i];
	}
	for (int i=n; i>n/2; i--) ans += dp[n][i];
	cout << fixed << setprecision(9) << ans << endl;
}
### 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;

int n;
double ans;

int main() {
	cin >> n;
	double p[n + 1], dp[n + 1][n + 1];
	dp[0][0] = 1;
	for (int i=1; i<=n; i++) {
		cin >> p[i];
		dp[i][0] = dp[i-1][0] * (1 - p[i]);
		for (int j=1; j<i; j++) {
			dp[i][j] = dp[i-1][j-1] * p[i] + dp[i-1][j] * (1 - p[i]);
		}
		dp[i][i] = dp[i-1][i-1] * p[i];
	}
	for (int i=n; i>n/2; i--) ans += dp[n][i];
	cout << fixed << setprecision(9) << ans << endl;
}
```