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

// --------------------------- Defines -------------------------------------  //
template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { return os << '(' << p.first << ", " << p.second << ')'; }
template<
    typename T_container, typename T = typename enable_if<!is_same<T_container, string>::value, typename T_container::value_type>::type
> ostream& operator<<(ostream &os, const T_container &v) { os << '{'; string sep; for (const T &x : v) os << sep << x, sep = ", "; return os << '}'; }
void dbg_out() { cerr << endl; }
template<typename Head, typename... Tail> void dbg_out(Head H, Tail... T) { cerr << ' ' << H; dbg_out(T...); }

#ifdef FILE_STREAM
streambuf *cin_backup = cin.rdbuf(), *cout_backup = cout.rdbuf();
ifstream f_in("name.in");
ofstream f_out("name.out");
#endif // FILE_STREAM

#ifdef DEBUG
    #define dbg(...) cerr << "(" << #__VA_ARGS__ << "):", dbg_out(__VA_ARGS__)
#else
    #define dbg(...)
#endif

#define ll long long
#define fill_array(x, y) memset(x, y, sizeof(x))
#define out_range_map(x,y) ((x) < 0 || (y) < 0 || (x) == N || (y) == M)
#define sza(x) ((int)x.size())
#define all(x) (x).begin(), (x).end()
#define fori(i, N) for(int i = 0; i < N; i++)
#define left(x) (x << 1 | 1)
#define right(x) (left(x) + 1)
#define half (l + r) / 2
// --------------------------- /Defines/ ------------------------------------ //

// --------------------------- Constants ----------------------------------- //
const ll LN = (ll)(5e3 + 5), 
         LM = (ll)(1e2 + 5), 
         INF = (ll)(1e15), 
         MOD = (ll)(1e9);
ll N, M, K, T, Q;
// -------------------------- /Constants/ ---------------------------------- //
vector < double > prob;
double memo[LN][LN];

double dp(int i, int k)
{
    if(memo[i][k] != -1)
        return memo[i][k];
    if(k > i + 1)
        return memo[i][k] = 0;
    if (k == 0)
        return memo[i][k] = (i ? dp(i-1,0)*(1-prob[i]) : (1-prob[0]));
    if (i == 0)
        return memo[0][1] = prob[0];
    
    return memo[i][k] = dp(i - 1, k)*(1-prob[i]) + dp(i - 1, k - 1) * prob[i];
}

// ------------------------- Your code ---------------------------------- //
void Solve() {
	
    cin >> N >> K;

    prob.resize(N);

    for (int i = 0; i < N; i++)
        cin >> prob[i];

    sort(all(prob), greater<double>());

    for (int i = 0; i <= N; i++)
        for (int j = 0; j <= N; j++)
            memo[i][j] = -1;

    double ans = -1.0, tans;
    for (int i = K-1; i < N; i++)
    {
        tans = 0.0;
        for (int k = (K + i + 2)/2; k <= i + 1; k++)
            tans += dp(i,k);
        ans = max(ans, tans);
    }

    cout << ans << '\n';
    
}
// --------------------------------------------------------------------- //

int main() {
    #ifdef DEBUG
        freopen("input.txt","w",stdin);
        freopen("output.txt","w",stdout);
    #else
        //ios_base::sync_with_stdio(0); cin.tie(0);
    #endif

    int Tc = 1;
    // cin >> Tc;

    fori(i, Tc)
        Solve();

    return 0;
}