#include<cstdio>

#define ran 3033

int n;
double dp[ran][ran];

int main() {
	scanf("%d", &n);
	dp[0][0] = 1.;
	for(int i=1;i<=n;i++) {
		double p;
		scanf("%lf", &p);
		for(int j=0;j<i;j++) {
			dp[i][j] += dp[i-1][j] * (1-p);
			dp[i][j+1] += dp[i-1][j] * p;
		}
	}
	double ans = 0;
	for(int i=n;i>n-i;i--)
		ans += dp[n][i];
	printf("%.10lf\n", ans);
	
	return 0;
}
### Prompt

Please formulate 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<cstdio>

#define ran 3033

int n;
double dp[ran][ran];

int main() {
	scanf("%d", &n);
	dp[0][0] = 1.;
	for(int i=1;i<=n;i++) {
		double p;
		scanf("%lf", &p);
		for(int j=0;j<i;j++) {
			dp[i][j] += dp[i-1][j] * (1-p);
			dp[i][j+1] += dp[i-1][j] * p;
		}
	}
	double ans = 0;
	for(int i=n;i>n-i;i--)
		ans += dp[n][i];
	printf("%.10lf\n", ans);
	
	return 0;
}
```