#include <bits/stdc++.h>
#define ll long long

using namespace std;

double p[3030];
double Dp[3030][3030];

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

    for(ll i=0;i<n;i++) scanf("%lf",p+i);

    for(ll H=0;H<=n;H++)
        if(H>n-H)
            Dp[n][H] = 1.0;

    for(ll i=n-1;i>=0;i--)
        for(ll H=0;H<=i;H++)
            Dp[i][H] = p[i]*Dp[i+1][H+1] + (1.0-p[i])*Dp[i+1][H];

    printf("%.10f",Dp[0][0]);
}

### Prompt

Please provide a cpp coded solution 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>
#define ll long long

using namespace std;

double p[3030];
double Dp[3030][3030];

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

    for(ll i=0;i<n;i++) scanf("%lf",p+i);

    for(ll H=0;H<=n;H++)
        if(H>n-H)
            Dp[n][H] = 1.0;

    for(ll i=n-1;i>=0;i--)
        for(ll H=0;H<=i;H++)
            Dp[i][H] = p[i]*Dp[i+1][H+1] + (1.0-p[i])*Dp[i+1][H];

    printf("%.10f",Dp[0][0]);
}

```