#include <iostream>
using namespace std;

int main() {
	int n;
	cin >> n;
	double dp[3000];
	dp[0] = 1;
	for (int j = 1; j <= n; j++) dp[j] = 0;
	for (int i = 1; i <= n; i++) {
		double p;
		cin >> p;
		for (int j = n; j >= 0; j--) {
			dp[j] *= (1-p);
			if (j >= 1) dp[j] += dp[j-1] * p;
		}
	}
	double ans = 0;
	for (int j = n/2 + 1; j <= n; j++) ans += dp[j];
	printf("%.15lf\n", ans);
	return 0;
}
### Prompt

Generate 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 <iostream>
using namespace std;

int main() {
	int n;
	cin >> n;
	double dp[3000];
	dp[0] = 1;
	for (int j = 1; j <= n; j++) dp[j] = 0;
	for (int i = 1; i <= n; i++) {
		double p;
		cin >> p;
		for (int j = n; j >= 0; j--) {
			dp[j] *= (1-p);
			if (j >= 1) dp[j] += dp[j-1] * p;
		}
	}
	double ans = 0;
	for (int j = n/2 + 1; j <= n; j++) ans += dp[j];
	printf("%.15lf\n", ans);
	return 0;
}
```