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;
pair<int, int> result[1401][1401];
int main() {
  int h, w;
  cin >> h >> w;
  int ans = 0;
  pair<int, int> zero = make_pair(0, 0);
  for (int i = 0; i < h; i++) {
    for (int j = 0; j < w; j++) {
      result[i][j] = zero;
      int c;
      cin >> c;
      if (c == 1) continue;
      if (i == 0 && j == 0) {
        result[i][j] = make_pair(1, 1);
      } else if (i == 0 || result[i - 1][j] == zero) {
        result[i][j].first = 1;
        result[i][j].second = result[i][j - 1].second + 1;
      } else if (j == 0 || result[i][j - 1] == zero) {
        result[i][j].first = result[i - 1][j].first + 1;
        result[i][j].second = 1;
      } else if (result[i - 1][j - 1] == zero) {
        if (result[i - 1][j].first > result[i][j - 1].second) {
          result[i][j].first = result[i - 1][j].first + 1;
          result[i][j].second = 1;
        } else {
          result[i][j].first = 1;
          result[i][j].second = result[i][j - 1].second + 1;
        }
      } else {
        result[i][j].first =
            min(result[i - 1][j].first,
                min(result[i][j - 1].first, result[i - 1][j - 1].first)) +
            1;
        result[i][j].second =
            min(result[i - 1][j].second,
                min(result[i][j - 1].second, result[i - 1][j - 1].second)) +
            1;
      }
      ans = max(ans, result[i][j].first * result[i][j].second);
    }
  }
  cout << ans << endl;
  return 0;
}
