#include <bits/stdc++.h>
using namespace std;
int N, M;
int A[2505][2505];
int up[2505][2505];
int dn[2505][2505];
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cin >> N >> M;
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++) {
char c;
cin >> c;
A[i][j] = c - '0';
}
}
// Solution:
//
// The upper bound on height and width is the minimum length
// vertical and horizontal strip. Now, for a fixed width Y,
// we want to determine the maximum height X. Note that,
// if we have width Y, then for every maximal horizontal
// strip, height is bounded by the minimum (top - bottom)
// of the prefix and suffix of size Y.
//
// We only need to consider the prefix and suffix. Why?
// For, after a prefix of size Y, let's move it to the right.
// If the (top - bottom) decreased, that means that there is
// a vertical strip which is shifted downwards, which means
// it will occcupy a maximal horizontal strip -> we will compute
// it later.
//
// Time complexity: O(N M).
for (int j = 1; j <= M; j++) {
for (int i = 1; i <= N; i++) {
if (A[i][j]) {
up[i][j] = 1 + up[i - 1][j];
}
}
for (int i = N; i >= 1; i--) {
if (A[i][j]) {
dn[i][j] = 1 + dn[i + 1][j];
}
}
}
vector<int> ans(M + 2, N);
for (int rep = 0; rep < 1; rep++) { // for prefix, then suffix
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++) if (A[i][j]) {
ans[1] = min(ans[1], up[i][j] + dn[i][j] - 1);
}
for (int j = 1; j <= M; j++) if (A[i][j] && !A[i][j - 1]) {
int jj = j;
while (A[i][jj + 1]) jj += 1;
int len = jj - j + 1;
int minUp = N;
int minDn = N;
for (int k = 1; k <= len; k++) {
minUp = min(minUp, up[i][j + k - 1]);
minDn = min(minDn, dn[i][j + k - 1]);
ans[k] = min(ans[k], minUp + minDn - 1);
}
ans[len + 1] = 0;
}
}
for (int i = 1; i <= N; i++) {
for (int j = 1; j + j <= M; j++) {
swap(A[i][j], A[i][M - j + 1]);
swap(up[i][j], up[i][M - j + 1]);
swap(dn[i][j], dn[i][M - j + 1]);
}
}
}
int answer = 0;
for (int i = 1; i <= M; i++) {
ans[i] = min(ans[i], ans[i - 1]);
answer = max(answer, i * ans[i]);
}
cout << answer << '\n';
return 0;
}
