Problem:
Score : 100 points

Problem Statement

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}.

Sample Input 1

3
0.30 0.60 0.80

Sample Output 1

0.612

The probability of each case where we have more heads than tails is as follows:

The probability of having (Coin 1, Coin 2, Coin 3) = (Head, Head, Head) is 0.3 × 0.6 × 0.8 = 0.144;

The probability of having (Coin 1, Coin 2, Coin 3) = (Tail, Head, Head) is 0.7 × 0.6 × 0.8 = 0.336;

The probability of having (Coin 1, Coin 2, Coin 3) = (Head, Tail, Head) is 0.3 × 0.4 × 0.8 = 0.096;

The probability of having (Coin 1, Coin 2, Coin 3) = (Head, Head, Tail) is 0.3 × 0.6 × 0.2 = 0.036.

Thus, the probability of having more heads than tails is 0.144 + 0.336 + 0.096 + 0.036 = 0.612.

Sample Input 2

1
0.50

Sample Output 2

0.5

Outputs such as 0.500, 0.500000001 and 0.499999999 are also considered correct.

Sample Input 3

5
0.42 0.01 0.42 0.99 0.42

Sample Output 3

0.3821815872

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <cmath>
#include <bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>

using namespace std;


#define ld long double
#define F first
#define S second
//
#define int long long
#define ll long long
#define db(x)                cerr<<#x<<": "<<x<<endl
#define db2(x, y)             cerr<<#x<<": "<<x<<" | "<<#y<<": "<<y<<endl
#define db3(x, y, z)          cerr<<#x<<":" <<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl
#define db4(a, b, c, d)       cerr<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<endl
#define db5(a, b, c, d, e)    cerr<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<" | "<<#e<< ": "<<e<<endl
#define db6(a, b, c, d, e, f) cerr<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<" | "<<#e<< ": "<<e<<" | "<<#f<<": "<<f<<endl
//
#define mem(x,y) memset(x,y,sizeof(x))
#define bitcount    __builtin_popcount
//
#define pii pair<int, int>
#define all(v) v.begin(),v.end()
//
#define IOS ios_base::sync_with_stdio(false); cin.tie(NULL)
#define vi vector<int>
#define vvi vector< vector<int> >
#define si set<int>
#define pb push_back
#define lp(i,a,b) for(int i=a;i<b;i++)
#define rloop(i,a,b) for(int i=a;i>=b;i--)
#define mk make_pair
const int N=3e3+5;
const int mod=1e9+7;
const int mod1=998244353;
const int inf=1e16+7;
using namespace std;

double dp[N][N];
double pr[N];
int n;

double rec(int i,int h=0){
  //  db2(i,h);
    if(i==n){
        int  t=n-h;
        if(h>t) return  1.0;
        else return 0;
    }
    if(dp[i][h]!=0.0) return dp[i][h];
    double ans1=rec(i+1,h+1)*pr[i];
    double ans2=rec(i+1,h)*(1.0-pr[i]);
    return dp[i][h]=ans1+ans2;
    
}
void solve(){
  //  memset(dp,-1,sizeof(dp));
    
    cin>>n;
    for(int i=0;i<n;i++)
        cin>>pr[i];
   // db("F");
    cout<<fixed<<setprecision(15)<<rec(0,0);
    //db("DEW");
}


int32_t main() {
    IOS;
    
    int t;
    //cin>>t;
    t=1;
    for(int i=1;i<=t;i++){
       solve();
    }
   
}