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)

#include <bits/stdc++.h>

using namespace std;

//#define PRAGMA 448
#ifdef PRAGMA
#pragma GCC optimize("Ofast")
#pragma GCC optimize ("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#endif

#ifdef LUCARIO
#define deb(...) fprintf(stderr,__VA_ARGS__)
#define deb1(x) cerr << #x << " = " << x << endl
#else
#define deb(...) 0
#define deb1(x) 0
#endif

typedef int intt;
typedef pair<long long,long long> pll;
typedef pair<int,int> pii;

#define FIN ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define ll long long
#define ld long double
#define int long long
#define pb push_back
#define bs binary_search
#define up upper_bound
#define fs first 
#define sd second 
#define mp make_pair    
#define endl "\n"

#define fore(i,a,b) for(ll i = a; i < b; i++)
#define rfore(i, a, b) for(ll i = a - 1; i >= b; i--)
#define forev(a,b) fore(i,0,b)cout<<a[i]<<" "; 
#define ing(g,u,v)  g[u].pb(v); g[v].pb(u);
#define ingp(g,u,v,c) (g[u].pb({v,c});g[v].pb({u,c});)
#define mm(a,b) memset(a , b , sizeof a) 
#define ALL(a) a.begin(),a.end() 
#define SZ(n) ((ll) (n).size())
#define sfind(s,t) (s.find(t) != string::npos)

#define infll (ll)1e15
#define infi  (int)(1e9+1)
#define PI 3.14159265359 
#define mod (ll)(1e9+7)
#define eps  1e-6

ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b,a%b); } 
ll lcm(ll a, ll b) { return a * (b / gcd(a, b)); }


int n,m;
ld a[1001];
ld dp[3001][3001];
double res;

ld solve(int i, int j){
    if(i == n)return (j > n/2);
    ld &r = dp[i][j];
    if(r >= 0)return r;
    r = 0;
    r += solve(i+1,j+1)*a[i]+solve(i+1,j)*(1-a[i]);
    return r;
}


int32_t main(int32_t argc, char const *argv[]){
    FIN;
    fore(i,0,3001){
        fore(j,0,3001){
            dp[i][j] =-1;
        }
    }
    cin>>n;
    fore(i,0,n)cin>>a[i];

    cout<<setprecision(16)<<solve(0,0)<<endl;

    return 0;
}