#include <bits/stdc++.h>
using namespace std;
double dp[3001][3001],ans=0.0;
int n;

int main(){
	cin >> n;
	dp[0][0] = 1;
	for(int i = 1 ; i <= n ; i += 1){
		double a;
		cin >> a;
		for(int j = 0 ; j <= i ; j += 1){
			dp[i][j] = dp[i-1][j]*(1.0-a);
			if(j){
				dp[i][j] += dp[i-1][j-1]*a;
			}
		}
	}
	for(int i = n/2+1 ; i <= n ; i += 1){
		ans += dp[n][i];
	}
	cout << setprecision(9) << fixed;
	cout << ans << endl;
}
### 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 <bits/stdc++.h>
using namespace std;
double dp[3001][3001],ans=0.0;
int n;

int main(){
	cin >> n;
	dp[0][0] = 1;
	for(int i = 1 ; i <= n ; i += 1){
		double a;
		cin >> a;
		for(int j = 0 ; j <= i ; j += 1){
			dp[i][j] = dp[i-1][j]*(1.0-a);
			if(j){
				dp[i][j] += dp[i-1][j-1]*a;
			}
		}
	}
	for(int i = n/2+1 ; i <= n ; i += 1){
		ans += dp[n][i];
	}
	cout << setprecision(9) << fixed;
	cout << ans << endl;
}
```