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: Accepted)

/**
 *   created: 18 August 2019  19:06:17
**/
#include<bits/stdc++.h>
using namespace std;

typedef vector<int> vi;
typedef pair<int , int> pii;
typedef vector< pair<int , int> > vii;
typedef long long ll;
typedef vector<long long> vl;
typedef pair<long long , long long> pll;
typedef vector< pair<long long , long long> > vll;

#define PB push_back
#define PPB pop_back
#define all(c) c.begin(),c.end()
#define rep(i,a,b) for(int i = (int)(a); i < (int)(b); i++)
#define rrep(i,a,b) for(int i = (int)(a-1); i>=(int)(b); i--)
#define INFLL 2000000000000000007
#define INF 2000000007
#define MOD 1000000007


#define TRACE

#ifdef TRACE
#define trace1(x)                cerr << #x << ": " << x << endl;
#define trace2(x, y)             cerr << #x << ": " << x << " | " << #y << ": " << y << endl;
#define trace3(x, y, z)          cerr << #x << ": " << x << " | " << #y << ": " << y << " | " << #z << ": " << z << endl;
#define trace4(a, b, c, d)       cerr << #a << ": " << a << " | " << #b << ": " << b << " | " << #c << ": " << c << " | " << #d << ": " << d << endl;
#define trace5(a, b, c, d, e)    cerr << #a << ": " << a << " | " << #b << ": " << b << " | " << #c << ": " << c << " | " << #d << ": " << d << " | " << #e << ": " << e << endl;
#define trace6(a, b, c, d, e, f) cerr << #a << ": " << a << " | " << #b << ": " << b << " | " << #c << ": " << c << " | " << #d << ": " << d << " | " << #e << ": " << e << " | " << #f << ": " << f << endl;

#else

#define trace1(x)
#define trace2(x, y)
#define trace3(x, y, z)
#define trace4(a, b, c, d)
#define trace5(a, b, c, d, e)
#define trace6(a, b, c, d, e, f)

#endif


void optimizeIO()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
}

//                                        //AUTHOR: Kaushalesh Shukla

//=========================================================================================================

int main()
{
    optimizeIO();
    int tt=1;
    // cin>>tt;
    while(tt--){
        int n;
        cin>>n;
        vector< double > h(n);
        rep(i,0,n)
            cin>>h[i];
        vector< vector<double> > mem(n+1,vector<double>(n,0));
        mem[0][0] = 1-h[0];
        mem[1][0] = h[0];
        rep(i,1,n){
            mem[0][i] = mem[0][i-1]*(1-h[i]);
            rep(j,1,n+1){
                mem[j][i] = mem[j-1][i-1]*h[i] + mem[j][i-1]*(1-h[i]);
            }
        }
        double res = 0;
        rep(i,n/2+1,n+1){
            // trace1(i);
            res += mem[i][n-1];
        }
        // rep(i,0,n){
        //     rep(j,0,n+1)
        //         cout<<mem[i][j]<<" ";
        //     cout<<endl;
        // }
        printf("%0.10f",res);
    }
    return 0;
}