#include<bits/stdc++.h>
using  namespace std;

long double n,a[3000],dp[3000][6001];

long double fun(int i,int k){
if(n-i+1<-k)return 0;
if(i>n)return 1;
if(dp[i][k+3000])return dp[i][k+3000];
return dp[i][k+3000]=a[i]*fun(i+1,k+1) + (1-a[i])*fun(i+1,k-1);
}

int main(){
    cout.precision(10);cout<<fixed;
cin>>n;

//memset(dp,lb,sizeof dp);
for(int i=1;i<=n;i++){
    cin>>a[i];
}
cout<<fun(1,0);
}

### 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;

long double n,a[3000],dp[3000][6001];

long double fun(int i,int k){
if(n-i+1<-k)return 0;
if(i>n)return 1;
if(dp[i][k+3000])return dp[i][k+3000];
return dp[i][k+3000]=a[i]*fun(i+1,k+1) + (1-a[i])*fun(i+1,k-1);
}

int main(){
    cout.precision(10);cout<<fixed;
cin>>n;

//memset(dp,lb,sizeof dp);
for(int i=1;i<=n;i++){
    cin>>a[i];
}
cout<<fun(1,0);
}

```