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
import java.io.*;
import java.util.*;
import java.math.*;
import java.text.DecimalFormat;
import java.text.NumberFormat;
 
public class Main {
      
    int mod = 1000000007;

    public class SegmentTree{

       int n;
       int[] tree;

       SegmentTree(int[] arr){

          this.n = arr.length; 
          this.tree = new int[2 * n];

          for(int i = n; i < 2 * n; i++){
              this.tree[i] = arr[i - n];
          }

          for(int i = n - 1; i >= 1; i--){
              this.tree[i] = this.tree[2 * i] & this.tree[2 * i + 1];
          }

       }

       public int query(int l, int r){

          int res = (1 << 31) - 1;
          l = l + n;
          r = r + n;

          while(l < r){

              if(l % 2 == 1){
                  res &= tree[l]; 
                  l++;
              }

              if(r % 2 == 1){
                  r--;
                  res &= tree[r];
              }

              l /= 2;
              r /= 2;
          }

          return res;
       }
    }

    public void solve() throws IOException{
    
       int n = in.nextInt();
       int m = in.nextInt();
       int ans = 0;

       List<String> mat = new ArrayList<>();
       for(int i = 0; i < n; i++){
           String str = in.next();
           mat.add(str);
       }
        
       int[] cum = new int[n];
       int[] arr = new int[n];

       for(int j = m - 1; j >= 0; j--){

          for(int i = 0; i < n; i++){

             int num = mat.get(i).charAt(j) - '0';
             if(num == 0){
                cum[i] = 0;
             }else{
                cum[i]++;
             }
               
             arr[i] = cum[i];
          }
         
          Arrays.sort(arr);
          int size = Integer.MAX_VALUE;
          int cnt = 0;

          for(int i = n - 1 ; i >= 0; i--){
             int num = arr[i];
             size = Math.min(size, num);
             cnt++;
             ans = Math.max(ans, size * cnt);
          }

       }

      
       out.println(ans);
       return;
    }



    FastScanner in;
    PrintWriter out;
    
    static class FastScanner {
 
        BufferedReader br;
        StringTokenizer st;
 
        FastScanner(InputStream in) {
            br = new BufferedReader(new InputStreamReader(in));
            st = null;
        }
 
        String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }
 
        String nextLine() throws IOException {
            if (st == null || !st.hasMoreTokens())
                return br.readLine();
            StringBuilder result = new StringBuilder(st.nextToken());
            while (st.hasMoreTokens()) {
                result.append(" ");
                result.append(st.nextToken());
            }
            return result.toString();
        }
 
        int nextInt() throws IOException {
            return Integer.parseInt(next());
        }
 
        long nextLong() throws IOException {
            return Long.parseLong(next());
        }
 
        double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }
 
    }
 
    void run() throws IOException {
        in = new FastScanner(System.in);
        out = new PrintWriter(System.out, false);
        solve();
        out.close();
    }
 
    public static void main(String[] args) throws IOException{
        new Main().run();
    }
 
    public void printArr(int[] arr){
        for(int i = 0; i < arr.length; i++){
            out.print(arr[i] + " ");
        }
        out.println();
    }
 
    public long gcd(long a, long b){
        if(a == 0) return b;
        return gcd(b % a, a);
    }

    public boolean isPrime(long num){

        if(num == 0 || num == 1){
            return false;
        }

        for(int i = 2; i * i <= num; i++){
            if(num % i == 0){
                return false;
            }
        }

        return true;
    }

    public class Pair<A, B>{
        public A x; 
        public B y;

        Pair(A x, B y){
            this.x = x;
            this.y = y;
        }

        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;
            Pair<?, ?> pair = (Pair<?, ?>) o;
            if (!x.equals(pair.x)) return false;
            return y.equals(pair.y);
        }

        @Override
        public int hashCode() {
            int result = x.hashCode();
            result = 31 * result + y.hashCode();
            return result;
        }

    }

    class Tuple{
        int x; int y; int z;
        Tuple(int ix, int iy, int iz){
            x = ix;
            y = iy;
            z = iz;
        }
    }
}
