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;
typedef long long ll;
#define fr(i,a,b) for (int i=(a); i<(b);i++)
#define frr(i,a,b) for(int i=(a-1);i>=(b);i--)
#define vi vector<int>
#define de deque<int>
#define del deque<ll>
#define vl vector<ll>
#define pb push_back
#define ppb pop_back()
#define pf push_front
#define ppf pop_front()
#define mp map<ll,ll>
#define rev(a) reverse(a.begin(),a.end())
#define srt(a) sort(a.begin(),a.end())
#define rsrt(a) sort(a.begin(),a.end(),greater<int>())
#define ln(a) a.length()
#define sz(a) a.size()
ll M=1e9+7;
 int n;double a[3001];
 double dp[3001][3001];
 int ready[3001][3001];
 double solve(int ind,int num)
 {
 	    if(ready[ind][num])return dp[ind][num];
      if(ind==0){return 0.0;}
      if(num==0)return (1-a[ind])*solve(ind-1,num);
         
       ready[ind][num]=1;
       return dp[ind][num]=a[ind]*solve(ind-1,num-1)+(1.0-a[ind])*solve(ind-1,num);


 }

int main()
{ios_base::sync_with_stdio(false);
    cin.tie(NULL);cout.tie(NULL);
    
      #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif
    cin>>n;
    fr(i,1,n+1)cin>>a[i];
    memset(ready,0,sizeof ready);
    fr(i,0,3001)fr(j,0,3001)dp[i][j]=0.0;
    ready[0][0]=1;dp[0][0]=1.0;

    double ans=0;
    fr(i,0,n+1)
    {
       if(i>n/2)ans+=solve(n,i);
    }

    cout<<setprecision(12)<<ans<<endl;
}