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)

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

#define _CRT_SECURE_NO_WARNINGS
#include <bits/stdc++.h>

using namespace std;

typedef pair <int, int> pii;
typedef pair <long long, long long> pll;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;

#define endl "\n"
#define mt make_tuple
#define mp make_pair

template <typename T1, typename T2>
bool umax(T1 &a, const T2&b) { return a < b ? a = b, 1 : 0;}
template <typename T1, typename T2> 
bool umin(T1 &a, const T2 &b) {return a > b ? a = b, 1 : 0;}
template <typename T>
T sqr(T a) {return a * a;}

mt19937 rng(20010709);
const int mod = 1000000007;
const int INF = 1000000001;

int n;
ld dp[3000][6005], p[3000];
ld result;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    cin >> n;

    for (int i = 1; i <= n; i++)
        cin >> p[i];

    dp[0][0 + n] = 1;

    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n * 2; j++) {
            dp[i][j + 1] += dp[i - 1][j] * p[i];
            dp[i][j - 1] += dp[i - 1][j] * (1.0 - p[i]);
        }
    }

    for (int i = 1; i <= n; i++)
        result += dp[n][i + n];

    cout << fixed << setprecision(15) << result << endl;

    return 0;
}