#include "bits/stdc++.h"
#include <array>
using namespace std;


int n, m;

double dp[3000] = {};

int main(){
	int n; cin >> n;

	dp[0] = 1.0;

	for(int i = 0; i < n; i++){
		double p; cin >> p;
		for(int j = i; j >= 0; j--){
			dp[j+1] += dp[j] * p;
			dp[j] *= (1.0 - p);
		}
	}

	double ans = 0.0;
	for(int i = n/2+1; i <= n; i++){
		ans += dp[i];
	}

	cout << setprecision(10) << ans << endl;

    return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

### Response

```cpp
#include "bits/stdc++.h"
#include <array>
using namespace std;


int n, m;

double dp[3000] = {};

int main(){
	int n; cin >> n;

	dp[0] = 1.0;

	for(int i = 0; i < n; i++){
		double p; cin >> p;
		for(int j = i; j >= 0; j--){
			dp[j+1] += dp[j] * p;
			dp[j] *= (1.0 - p);
		}
	}

	double ans = 0.0;
	for(int i = n/2+1; i <= n; i++){
		ans += dp[i];
	}

	cout << setprecision(10) << ans << endl;

    return 0;
}

```