I

Problem Statement

Let $\mathrm{NN}$ be a positive odd number.

There are $\mathrm{NN}$ coins, numbered $\mathrm{1,2,\dots ,N1,2,\dots ,N}$. For each $\mathrm{ii}$ ($\mathrm{1\le i\le N1\le i\le N}$), when Coin $\mathrm{ii}$ is tossed, it comes up heads with probability ${\mathrm{pipi}}^{}$ and tails with probability $\mathrm{1-pi1-pi}$.

Taro has tossed all the $\mathrm{NN}$ coins. Find the probability of having more heads than tails.

Constraints

• $\mathrm{NN}$ is an odd number.
• $\mathrm{1\le N\le 29991\le N\le 2999}$
• ${\mathrm{pipi}}^{}$ is a real number and has two decimal places.
• $\mathrm{0<pi<10<pi<1}$

---

Input

Input is given from Standard Input in the following format:

$\mathrm{NN}$
${\mathrm{p1p1}}^{}$ ${\mathrm{p2p2}}^{}$ $\dots \dots$ ${\mathrm{pNpN}}^{}$

Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than ${\mathrm{10-910-9}}^{}$.

---

Sample Input 1 Copy

Copy

3
0.30 0.60 0.80

Sample Output 1 Copy

Copy

0.612

The probability of each case where we have more heads than tails is as follows:

• The probability of having $(Coin1,Coin2,Coin3)=(Head,Head,Head)(Coin1,Coin2,Coin3)=(Head,Head,Head)$ is $\mathrm{0.3\times 0.6\times 0.8=0.1440.3\times 0.6\times 0.8=0.144}$;
• The probability of having $(Coin1,Coin2,Coin3)=(Tail,Head,Head)(Coin1,Coin2,Coin3)=(Tail,Head,Head)$ is $\mathrm{0.7\times 0.6\times 0.8=0.3360.7\times 0.6\times 0.8=0.336}$;
• The probability of having $(Coin1,Coin2,Coin3)=(Head,Tail,Head)(Coin1,Coin2,Coin3)=(Head,Tail,Head)$ is $\mathrm{0.3\times 0.4\times 0.8=0.0960.3\times 0.4\times 0.8=0.096}$;
• The probability of having $(Coin1,Coin2,Coin3)=(Head,Head,Tail)(Coin1,Coin2,Coin3)=(Head,Head,Tail)$ is $\mathrm{0.3\times 0.6\times 0.2=0.0360.3\times 0.6\times 0.2=0.036}$.

Thus, the probability of having more heads than tails is $\mathrm{0.144+0.336+0.096+0.036=0.6120.144+0.336+0.096+0.036=0.612}$.

---

Sample Input 2 Copy

Copy

1
0.50

Sample Output 2 Copy

Copy

0.5

Outputs such as 0.500, 0.500000001 and 0.499999999 are also considered correct.

---

Sample Input 3 Copy

Copy

5
0.42 0.01 0.42 0.99 0.42

Sample Output 3 Copy

Copy

0.3821815872

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <iomanip>
#include <deque>
#include <bitset>
//#include <unordered_set>
//#include <unordered_map>
#define ll              long long
#define pii             pair<int, int>
#define rep(i,a,b)      for(int  i=a;i<=b;i++)
#define dec(i,a,b)      for(int  i=a;i>=b;i--)
using namespace std;
int dir[4][2] = { { 1,0 },{ 0,1 } ,{ 0,-1 },{ -1,0 } };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
const double pi = 3.14159265358979323846;
const double eps = 1e-6;
const int mod = 1e9 + 7;
const int N = 3e3 + 5;
//if(x<0 || x>=r || y<0 || y>=c)

inline ll read()
{
    ll x = 0; bool f = true; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? x : -x;
}
ll gcd(ll m, ll n)
{
    return n == 0 ? m : gcd(n, m % n);
}
ll lcm(ll m, ll n)
{
    return m * n / gcd(m, n);
}
bool prime(int x) {
    if (x < 2) return false;
    for (int i = 2; i * i <= x; ++i) {
        if (x % i == 0) return false;
    }
    return true;
}
ll qpow(ll m, ll k, ll mod)
{
    ll res = 1, t = m;
    while (k)
    {
        if (k & 1)
            res = res * t % mod;
        t = t * t % mod;
        k >>= 1;
    }
    return res;
}
double dp[N][N];
double a[N],ans;
int main()
{
    ll n;
    cin >> n;
    rep(i, 1, n)
        cin >> a[i];
    dp[0][0] = 1;
    rep(i, 1, n)
    {
        rep(j, 0, i)
        {
            int x = j, y = i - j;
            if (x)
                dp[x][y] += dp[x - 1][y] * a[i];
            if (y)
                dp[x][y] += dp[x][y - 1] * (1 - a[i]);
        }
    }
    int l = n, r = 0;
    while (l > r)
    {
        ans += dp[l][r];
        l--, r++;
    }
    printf("%.10f
", ans);
    return 0;
}