#include<bits/stdc++.h>
using namespace std;
long long const MAX=1e9+7;
double dp[3002];
int main()
{long long i,j,n,r,k,p,w,y,sum,m;
double x;
x=1;
cin>>n;
double a[n];
k=n/2;
for(i=0;i<n;i++)
{scanf("%lf",&a[i]);}
dp[0]=x;
for(i=1;i<=n;i++)
{for(j=i;j>=0;j--)
{ dp[j]=dp[j]*(x-a[i-1]);
if(j>0)
dp[j]=dp[j]+dp[j-1]*a[i-1];
}}
x=0;
for(i=k+1;i<=n;i++)
{x=x+dp[i];}
printf("%.10lf",x);
return 0;}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
long long const MAX=1e9+7;
double dp[3002];
int main()
{long long i,j,n,r,k,p,w,y,sum,m;
double x;
x=1;
cin>>n;
double a[n];
k=n/2;
for(i=0;i<n;i++)
{scanf("%lf",&a[i]);}
dp[0]=x;
for(i=1;i<=n;i++)
{for(j=i;j>=0;j--)
{ dp[j]=dp[j]*(x-a[i-1]);
if(j>0)
dp[j]=dp[j]+dp[j-1]*a[i-1];
}}
x=0;
for(i=k+1;i<=n;i++)
{x=x+dp[i];}
printf("%.10lf",x);
return 0;}

```