#include <fstream>
#include <iomanip>
#include <deque>

#define lint long long int
using namespace std;

const int NMAX = 155;

int n, m, r, c;
int mat[2 * NMAX][2 * NMAX];

lint s_part[2 * NMAX][2 * NMAX];
deque <int> coada;

inline lint sum (int l1, int l2, int c, int x) {
    return s_part[l2][c] - s_part[l1 - 1][c] - c * (l2 - l1 + 1ll) * x;
}

bool exists (int x) {
    int j, k;

    for (int i = 1; i <= 2 * n; ++ i) {
        for (j = i + r - 1; j <= 2 * n && j <= i + n - 1; ++ j) {
            coada.clear();

            for (k = 1; k <= 2 * m; ++ k) {
                while (!coada.empty() && coada.front() < k - n)
                    coada.pop_front();

                if (k >= c) {
                    while (!coada.empty() && sum(i, j, coada.back(), x) >= sum(i, j, k - c, x))
                        coada.pop_back();
                    coada.push_back(k - c);


                    if (sum(i, j, k, x) - sum(i, j, coada.front(), x) >= 0)
                        return true;
                }
            }
        }
    }

    return false;
}

int main()
{
    ifstream cin("balans.in");
    ofstream cout("balans.out");

    cin >> n >> m >> r >> c;

    if (!r) r = 1;
    if (!c) c = 1;

    int j;
    for (int i = 1; i <= n; ++ i)
        for (j = 1; j <= m; ++ j)
            cin >> mat[i][j];

    for (int i = 1; i <= n; ++ i)
        for (j = 1; j <= m; ++ j)
            mat[i][j] *= 1000;

    for (int i = 1; i <= 2 * n; ++ i)
        for (j = 1; j <= 2 * m; ++ j) {
            mat[i][j] = mat[(i - 1) % n + 1][(j - 1) % m + 1];
            s_part[i][j] = mat[i][j] + s_part[i][j - 1] + s_part[i - 1][j] - s_part[i - 1][j - 1];
        }

    /*int st = 1;
    int dr = 100000000;
    int ans = 0;

    int mijl;
    while (st <= dr) {
        mijl = (st + dr) >> 1;

        if (exists(mijl)) {
            ans = mijl;
            st = mijl + 1;
        }
        else
            dr = mijl - 1;
    }*/

    int ans = 0;
    for (int steps = 26; steps >= 0; -- steps)
        if (exists(ans | (1 << steps)))
            ans |= (1 << steps);

    cout << fixed << setprecision(3) << ans / 1000 << '.' << setw(3) << setfill('0') << ans % 1000 << '\n';

    cin.close();
    cout.close();
    return 0;
}