#include <bits/stdc++.h>

using namespace std;

int n;
double p, ans;
double dp[3010];


int main() {
	cin >> n; 
	dp[0] = 1;
	for (int i = 1; i <= n; i++) {
		cin >> p;
		for (int coins = i; coins >= 0; coins--) {
			dp[coins] = (coins == 0 ? 0 : dp[coins - 1] * p) + dp[coins] * (1 - p);
		}
	}
	cout.precision(20);
	for (int i = n / 2 + 1; i <= n; i++) 
		ans += dp[i];           
	cout << ans;
	return 0;
}
### Prompt

Please create a solution in cpp 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 <bits/stdc++.h>

using namespace std;

int n;
double p, ans;
double dp[3010];


int main() {
	cin >> n; 
	dp[0] = 1;
	for (int i = 1; i <= n; i++) {
		cin >> p;
		for (int coins = i; coins >= 0; coins--) {
			dp[coins] = (coins == 0 ? 0 : dp[coins - 1] * p) + dp[coins] * (1 - p);
		}
	}
	cout.precision(20);
	for (int i = n / 2 + 1; i <= n; i++) 
		ans += dp[i];           
	cout << ans;
	return 0;
}
```