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#include <bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>

//#pragma GCC optimize("O0")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,tune=native")
#pragma GCC optimize("unroll-loops")

using namespace std;
//using namespace __gnu_pbds;

#define pb push_back
#define F first
#define S second
#define ll long long
#define ld double
#define ull unsigned long long
#define endl '\n'

const ll N = 5e4 + 36;
const ll M = 2e3 + 36;
const ll INF = 1e9 + 7;
const ll MOD = 1e9 + 7;
const ll MOD1 = 888888901;
const ll MOD2 = 999988901;
const ll X[8] = {1, -1, 2, 2, -2, -2, 1, -1};
const ll Y[8] = {2, 2, 1, -1, 1, -1, -2, -2};
const ld PI = 3.14159265358979323846;
const ld EPS = 1e-10;

//tree < pair < string, int >, null_type, less < pair < string, int > >, rb_tree_tag, tree_order_statistics_node_update > a;

//mt19937 gen(chrono::system_clock::now().time_since_epoch().count());
mt19937 gen(19937);

signed main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
#ifdef LOCAL
    freopen("input.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#else
//    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);
#endif // LOCAL
    int n, t;
    cin >> n >> t;
    vector<ld> a(n);
    vector<vector<ld>> dp(N, vector<ld>(2, 0));
    int sqrt_N = 1200;

    for (int i(0); i < n; ++i) {
        cin >> a[i];
    }
    sort(a.rbegin(), a.rend());

    dp[0][0] = 1;
    ld ans = 0;
    int l = 0, r = 1, mid = -1;
    for (int i(0); i < n; ++i) {
        ld hig = 0, cur = 0;
        mid = l - 1;
        int lol = -1;
        if (i + 1 >= t) {
            lol = (i + 1 - t) / 2;
        }

        for (int j = l; j < r; ++j) {
            dp[j][(i % 2) ^ 1] += dp[j][i % 2] * a[i];
            dp[j + 1][(i % 2) ^ 1] += dp[j][i % 2] * (1.0 - a[i]);
            if (dp[j][(i % 2) ^ 1] > hig) {
                hig = dp[j][(i % 2) ^ 1];
                mid = j;
            }
            if (dp[j + 1][(i % 2) ^ 1] > hig) {
                hig = dp[j + 1][(i % 2) ^ 1];
                mid = j + 1;
            }
            dp[j][i % 2] = 0;
        }

        for (int j = l; j < r + 4; ++j) {
            if (max(0, mid - sqrt_N) > j || min(mid + sqrt_N, n + 1) <= j) {
                dp[j][(i % 2) ^ 1] = 0;
            } else if (j <= lol) {
                cur += dp[j][(i % 2) ^ 1];
            }
        }

        l = max(0, mid - sqrt_N), r = min(mid + sqrt_N, n + 1);
        ans = max(ans, cur);
    }

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

    return 0;
}