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: Runtime Error)

#pragma warning(disable:4996)
#pragma comment(linker, "/STACK:336777216")
#pragma GCC optimize("Ofast")

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

using namespace std;
using namespace __gnu_pbds;

typedef long long ll;
#define int long long int
typedef complex<long double> cd;
const long double pi=acos(-1);
typedef double db;
typedef long double ldb;
typedef pair <int, int> pii;
typedef pair <db, db> pdd;
typedef vector<int> vi;
typedef vector<vector<int> > matrix;

#define m1 make_pair
#define pb push_back
#define flush fflush(stdout)
#define IOS std::ios::sync_with_stdio(false); cin.tie(NULL);cout.tie(NULL);
#define reset(x,v) memset(x,v,sizeof(x))

#define fi first
#define se second
#define endl "\n"
#define debug(x) (cerr << #x << ": " << x << "\n")
#define setbits(x) __builtin_popcount(x)
#define setbitsll(x) __builtin_popcountll(x)
#define all(x) x.begin(),x.end()
#define pii pair<int,int>

inline ll gcd(ll a,ll b){if(a==0) return b;return gcd(b%a,a);}
inline ll power(ll a,ll n,ll m){if(n==0)return 1;ll p=power(a,n/2,m);p=(p%m*p%m)%m;if(n%2) return (p%m*a%m)%m;else return p;}
const double EPS = 1e-9;
const ll MOD =998244353;
const ll hell=1000000007;
const int INF = 1e18;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
template<typename T>
using ordered_set=tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

auto seed = chrono::high_resolution_clock::now().time_since_epoch().count();
std::mt19937 rng(seed);

double startTime;
double getCurrentTime() {
    return ((double)clock() - startTime) / CLOCKS_PER_SEC;
}
const int dx[] = { -1, 0, 1, 0 };
const int dy[] = { 0, 1, 0, -1 };

const int N=100005;

ldb dp[3000][300];

void solve()
{
   
    int n;
    cin>>n;
    ldb p[n+1];
    for(int i=1;i<=n;i++)
        cin>>p[i];
    dp[0][0]=1;
    for(int i=1;i<=n;i++)
    {
        for(int j=0;j<=i;j++)
        {
            if(!j)
                dp[i][j]=(1-p[i])*dp[i-1][j];
            else
                dp[i][j]=(1-p[i])*dp[i-1][j]+p[i]*dp[i-1][j-1];
        }
    }
    cout<<setprecision(20);
    ldb ans=0;
    for(int i=(n+1)/2;i<=n;i++)
        ans=ans+dp[n][i];
    cout<<ans<<endl;
    
    
    
    
    
}


int32_t main()
{
    IOS;
    int t;
    t=1;
    //cin>>t;
    while(t--)
    {
        solve();
    }
    return 0;
}