/*
 *使用堆辅助求解矩阵的最大面积
 *本题是求次大面积，故在原基础上有所修改
 */
#include <iostream>
#include <stack>

using namespace std;

const int N = 1e3 + 10;

int n, m;
char g[N][N];
int s[N][N];

int max1, max2;
int get(int);
int calc(int, int);

//关键算法
void bfs()
{
    for (int i = 1; i <= n; i ++ )
    {
        stack<int> st;
        st.push(0);
        for (int j = 1; j <= m + 1; j ++ )
        {
            while (s[i][st.top()] > s[i][j])
            {
                int index = st.top();
                st.pop();
                calc(j - 1 - st.top(), s[i][index]);//宽 * 高
                /*
                 *其中在st 到 j - 1之间的高度一定满足 大于s[i][j]
                 */
            }
            st.push(j);
        }
    }
}

int main()
{
    cin >> n >> m;
    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= m; j ++ )
        {
            cin >> g[i][j];
            if (g[i][j] == '1') 
                s[i][j] = s[i - 1][j] + 1;//计数高度，辅助搜索
        }

    bfs();

    cout << max2 << '\n';

    return 0;
} 


void get(int t)
{
    if (t >= max1) max2 = max1, max1 = t;
    else if (t > max2) max2 = t;
}

void calc(int x, int y)
{
    get(x * y);
    get((x - 1) * y);
    get(x * (y - 1));
}