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>
using namespace std;
static const int MAX_NUM = 140;
int LargestRectangle(bool area[][MAX_NUM + 1], int height, int width) {
  int i, j;
  int rect[MAX_NUM + 1][MAX_NUM + 1] = {0};
  int maximum = 0;
  for (i = 1; i <= width; i++) {
    for (j = 1; j <= height; j++) {
      if (area[j][i] == false) {
        rect[j][i] = rect[j - 1][i] + 1;
      }
    }
  }
  for (i = height; i >= 1; i--) {
    stack<pair<int, int> > stRect;
    pair<int, int> topelem;
    for (j = 1; j <= width; j++) {
      if (stRect.empty() == false) {
        topelem = stRect.top();
        if (topelem.first > rect[i][j]) {
          while ((topelem.first > rect[i][j]) && (stRect.empty() == false)) {
            stRect.pop();
            maximum = max(maximum, (j - topelem.second) * topelem.first);
          }
        } else if (topelem.first == rect[i][j]) {
          continue;
        }
      }
      stRect.push(make_pair(rect[i][j], j));
    }
    while (stRect.empty() == false) {
      topelem = stRect.top();
      stRect.pop();
      maximum = max(maximum, (width - topelem.second) * topelem.first);
    }
  }
  return maximum;
}
int main(void) {
  int i, j;
  int ret;
  int height, width;
  bool area[MAX_NUM + 1][MAX_NUM + 1] = {0};
  scanf("%d %d", &height, &width);
  for (i = 1; i <= height; i++) {
    for (j = 1; j <= width; j++) {
      scanf("%d", &area[i][j]);
    }
  }
  ret = LargestRectangle(area, height, width);
  printf("%d\n", ret);
  return 0;
}
