#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n, m;
ll a, b;
cin >> n >> m >> a >> b;
vector<vector<ll>> A(n+1, vector<ll>(m+1, 0));
for(int i=1; i<=n; i++){
for(int j=1; j<=m; j++){
cin >> A[i][j];
}
}
// Prefix sum per kolom
vector<vector<ll>> pref(n+1, vector<ll>(m+1, 0));
for(int j=1; j<=m; j++){
for(int i=1; i<=n; i++){
pref[i][j] = pref[i-1][j] + A[i][j];
}
}
auto dist = [&](ll sum){
return llabs(a - sum) + llabs(b - sum);
};
ll ans = LLONG_MAX;
// Enumerasi baris atas i, baris bawah x
for(int i=1; i<=n; i++){
for(int x=i; x<=n; x++){
vector<ll> col(m+1, 0);
for(int j=1; j<=m; j++){
col[j] = pref[x][j] - pref[i-1][j];
}
// Prefix sum untuk array col
ll curr = 0;
set<ll> S;
S.insert(0); // prefix awal
for(int j=1; j<=m; j++){
curr += col[j];
// Kita butuh cari prefix p sehingga
// (curr - p) dekat dengan a dan b
for(ll target : {a, b, (a+b)/2}) {
ll want = curr - target;
auto it = S.lower_bound(want);
if(it != S.end()){
ll sumRect = curr - *it;
ans = min(ans, dist(sumRect));
}
if(it != S.begin()){
--it;
ll sumRect = curr - *it;
ans = min(ans, dist(sumRect));
}
}
S.insert(curr);
}
}
}
cout << ans << "\n";
return 0;
}
