#include <bits/stdc++.h>

using namespace std;

double mem[3000][3000];
int vis[3000][3000];
int n;
double arr[3000];

double go(int x,int y)
{
	if(x>=n) return (y>n-y);
	if(vis[x][y]) return mem[x][y];
	vis[x][y]=1;
	return mem[x][y]=(arr[x]*go(x+1,y+1)+(1-arr[x])*go(x+1,y));
}

int main()
{
	cin >> n;
	for(int i=0;i<n;i++)
		cin >> arr[i];
	cout << fixed << setprecision(15) << go(0,0) << endl;
}
### Prompt

Generate 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 <bits/stdc++.h>

using namespace std;

double mem[3000][3000];
int vis[3000][3000];
int n;
double arr[3000];

double go(int x,int y)
{
	if(x>=n) return (y>n-y);
	if(vis[x][y]) return mem[x][y];
	vis[x][y]=1;
	return mem[x][y]=(arr[x]*go(x+1,y+1)+(1-arr[x])*go(x+1,y));
}

int main()
{
	cin >> n;
	for(int i=0;i<n;i++)
		cin >> arr[i];
	cout << fixed << setprecision(15) << go(0,0) << endl;
}
```