#include <cstdio>
double dp[3000];
int main() {
	int n;
	scanf("%d", &n);
	dp[0] = 1;
	for (int i = 0; i < n; i++) {
		double p;
		scanf("%lf", &p);
		for (int j = i + 1; j > 0; j--) {
			dp[j] = dp[j-1] * p + dp[j] * (1 - p);
		}
		dp[0] = dp[0] * (1 - p);
	}
	double ans = 0;
	for (int i = 0; i <= n; i++) {
		if (i > n - i) {
			ans += dp[i];
		}
	}
	printf("%.12f\n", ans);
	return 0;
}

### 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 <cstdio>
double dp[3000];
int main() {
	int n;
	scanf("%d", &n);
	dp[0] = 1;
	for (int i = 0; i < n; i++) {
		double p;
		scanf("%lf", &p);
		for (int j = i + 1; j > 0; j--) {
			dp[j] = dp[j-1] * p + dp[j] * (1 - p);
		}
		dp[0] = dp[0] * (1 - p);
	}
	double ans = 0;
	for (int i = 0; i <= n; i++) {
		if (i > n - i) {
			ans += dp[i];
		}
	}
	printf("%.12f\n", ans);
	return 0;
}

```