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;
typedef pair<ll, ll> P;
const ll MOD = 1e9+7;
const ll INF = 1e18;
#define rep(i,m,n) for(ll i = (m); i < (n); i++)
#define rrep(i,m,n) for(ll i = (m); i >= (n); i--)
#define print(x) cout << (x) << endl;
#define printa(x,m,n) for(ll i = (m); i <= n; i++){cout << (x[i]) << " ";} cout<<endl;

int main(){
	cin.tie(0); ios::sync_with_stdio(false);
	
	ll n;
	cin >> n;
	double p[n+1];
	rep(i, 1, n+1){
		cin >> p[i];
	}
	
	double dp[n+1][n+1][n+1];
	dp[0][0][0] = 0;
	rep(i, 1, n+1){
		for(ll k = 0; k <= i; k++){
			//dp[i][k][i-k] = 0;
			double buf = 0;
			if(k > 0){
				buf += exp(dp[i-1][k-1][i-k]+log(p[i]));
			}
			if(k < i){
				buf += exp(dp[i-1][k][i-k-1]+log(1-p[i]));
			}
			dp[i][k][i-k] = log(buf);
		}
	}
	
	double ans = 0;
	for(ll k = 0; k <= n; k++){
		if(k > n-k){
			ans += exp(dp[n][k][n-k]);
		}
	}
	cout << setprecision(10) << ans << endl;
	return 0;
}