Given a matrix (H × W) which contains only 1 and 0, find the area of the largest rectangle which only contains 0s.

Constraints

* 1 ≤ H, W ≤ 1,400

Input


H W
c1,1 c1,2 ... c1,W
c2,1 c2,2 ... c2,W
:
cH,1 cH,2 ... cH,W


In the first line, two integers H and W separated by a space character are given. In the following H lines, ci,j, elements of the H × W matrix, are given.

Output

Print the area (the number of 0s) of the largest rectangle.

Example

Input

4 5
0 0 1 0 0
1 0 0 0 0
0 0 0 1 0
0 0 0 1 0


Output

6
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
int mp[1500][1500]{0};
int t[1500][1500]{0};
int calRectangle(int (&vec)[1500], int w, int time) {
  int result = 0;
  stack<int> stk;
  vec[w + 1] = 0;
  if (true) {
    for (int i = 1; i <= w; i++) cout << vec[i] << " ";
    cout << endl;
  }
  for (int i = 1; i <= w + 1; i++) {
    if (stk.empty() || vec[stk.top()] <= vec[i])
      stk.push(i);
    else {
      int temp = stk.top();
      stk.pop();
      result = max(result,
                   vec[temp] * (stk.empty() ? (i - 1) : (i - stk.top() - 1)));
      i--;
    }
  }
  return result;
}
int main() {
  int h, w, pt, res = 0;
  cin >> h >> w;
  for (int i = 1; i <= h; i++)
    for (int j = 1; j <= w; j++) cin >> mp[i][j];
  for (int j = 1; j <= w; j++) {
    pt = 1;
    for (int i = 1; i <= h; i++)
      if (!mp[i][j])
        t[i][j] = pt++;
      else {
        t[i][j] = 0;
        pt = 1;
      }
  }
  for (int i = 1; i <= h; i++) res = max(res, calRectangle(t[i], w, i));
  cout << res << endl;
  return 0;
}
