#include <bits/stdc++.h>
using namespace std;
int main() {
	int n;
	scanf("%d", &n);
	vector <double> dp(n + 1);
	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] = p * dp[j - 1] + (1 - p) * dp[j];
		}
	}
	double ans = 0;
	for(int i = n / 2 + 1; i <= n; i ++) {
		ans += dp[i];
	}
	printf("%.10lf", ans);
	return 0;
}

### 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
#include <bits/stdc++.h>
using namespace std;
int main() {
	int n;
	scanf("%d", &n);
	vector <double> dp(n + 1);
	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] = p * dp[j - 1] + (1 - p) * dp[j];
		}
	}
	double ans = 0;
	for(int i = n / 2 + 1; i <= n; i ++) {
		ans += dp[i];
	}
	printf("%.10lf", ans);
	return 0;
}

```