You are given a matrix consisting of digits zero and one, its size is n × m. You are allowed to rearrange its rows. What is the maximum area of the submatrix that only consists of ones and can be obtained in the given problem by the described operations?

Let's assume that the rows of matrix a are numbered from 1 to n from top to bottom and the columns are numbered from 1 to m from left to right. A matrix cell on the intersection of the i-th row and the j-th column can be represented as (i, j). Formally, a submatrix of matrix a is a group of four integers d, u, l, r (1 ≤ d ≤ u ≤ n; 1 ≤ l ≤ r ≤ m). We will assume that the submatrix contains cells (i, j) (d ≤ i ≤ u; l ≤ j ≤ r). The area of the submatrix is the number of cells it contains.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 5000). Next n lines contain m characters each — matrix a. Matrix a only contains characters: "0" and "1". Note that the elements of the matrix follow without any spaces in the lines.

Output

Print a single integer — the area of the maximum obtained submatrix. If we cannot obtain a matrix of numbers one, print 0.

Examples

Input

1 1
1


Output

1


Input

2 2
10
11


Output

2


Input

4 3
100
011
000
101


Output

2

/* 
Keep solving problems. 
*/
import java.util.*;
import java.io.*;

public class CF375B {

    BufferedReader br;
    PrintWriter out;
    StringTokenizer st;
    boolean eof;
    final long MOD = 1000L * 1000L * 1000L + 7;
    int[] csort;
    int[] hist;
    void solve() throws IOException {
        int n = nextInt();
        int m = nextInt();

        int[][] acc = new int[n][m];
        char[][] map = new char[n][m];

        csort = new int[m];
        hist = new int[1 + m];


        for(int i = 0; i < n; i++) {
            map[i] = nextString().toCharArray();
        }
        for(int i = 0; i < n; i++) {
            acc[i][0] = map[i][0] == '0' ? 0 : 1;
        }
        for(int j = 1; j < m; j++) {
            for(int i = 0; i < n; i++) {
                acc[i][j] = map[i][j] == '0' ? 0 : 1 + acc[i][j - 1];
            }
        }

        int[] arr = new int[n];
        int res = 0;
        for(int j = 0; j < m; j++) {
            for(int i = 0; i < n; i++) {
                arr[i] = acc[i][j];
            }
            countSort(arr);

            for(int i = 0; i < n; i++) {
                int tmp = arr[i] * (n - i);
                res = Math.max(res, tmp);
            }
        }
        out(res);
    }
    void countSort(int[] arr) {
        for(int i = 0; i < hist.length; i++) {
            hist[i] = 0;
        }
        for(int v : arr) {
            hist[v]++;
        }
        int idx = 0;
        for(int i = 0; i < hist.length; i++) {
            for(int j = 0; j < hist[i]; j++) {
                arr[idx + j] = i;
            }
            idx += hist[i];
        }
    }

    void shuffle(int[] a) {
        int n = a.length;
        for(int i = 0; i < n; i++) {
            int r = i + (int) (Math.random() * (n - i));
            int tmp = a[i];
            a[i] = a[r];
            a[r] = tmp;
        }
    }
    private void outln(Object o) {
        out.println(o);
    }
    private void out(Object o) {
        out.print(o);
    }
    public CF375B() throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        long start = System.currentTimeMillis();
        solve();

        long end = System.currentTimeMillis();
//        System.out.println("take: " + (end - start) / 1000.0);
        out.close();
    }
    public static void main(String[] args) throws IOException {
        new CF375B();
    }

    public long[] nextLongArr(int n) throws IOException{
        long[] res = new long[n];
        for(int i = 0; i < n; i++)
            res[i] = nextLong();
        return res;
    }
    public int[] nextIntArr(int n) throws IOException {
        int[] res = new int[n];
        for(int i = 0; i < n; i++)
            res[i] = nextInt();
        return res;
    }
    public String nextToken() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                eof = true;
                return null;
            }
        }
        return st.nextToken();
    }
    public String nextString() {
        try {
            return br.readLine();
        } catch (IOException e) {
            eof = true;
            return null;
        }
    }
    public int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }
    public long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }
    public double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }
}

