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

double dp[3010];

int main(){
	int n;
	scanf("%d", &n);
	double p;
	dp[0] = 1;
	for(int i=0; i < n; i++) {
		scanf("%lf", &p);
		for(int j=i+1; j >= 0; j--)
			dp[j] = (j == 0 ? 0 : dp[j-1]*p) + dp[j]*(1-p);
	}
	double ans = 0;
	for(int i=0; i <= n; i++)
		if(i > n-i) ans += dp[i];
	printf("%.12lf\n", ans);
}

### Prompt

Generate a CPP solution to the following problem:

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>
using namespace std;

double dp[3010];

int main(){
	int n;
	scanf("%d", &n);
	double p;
	dp[0] = 1;
	for(int i=0; i < n; i++) {
		scanf("%lf", &p);
		for(int j=i+1; j >= 0; j--)
			dp[j] = (j == 0 ? 0 : dp[j-1]*p) + dp[j]*(1-p);
	}
	double ans = 0;
	for(int i=0; i <= n; i++)
		if(i > n-i) ans += dp[i];
	printf("%.12lf\n", ans);
}

```