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;
int H, W;
int R[1405][1405];
int rec(int f1, int c1, int f2, int c2) {
  if (f1 < 0 || f1 > H - 1 || f2 < 0 || f2 > H - 1 || c1 < 0 || c1 > W - 1 ||
      c2 < 0 || c2 > W - 1 || f1 > f2 || c1 > c2)
    return -1;
  if (f1 == f2 && c1 == c2) return 1 - R[f1][c1];
  for (int i = f1; i <= f2; i++)
    for (int j = c1; j <= c2; j++)
      if (R[i][j]) {
        return max(rec(f1, c1, i - 1, c2),
                   max(rec(i + 1, c1, f2, c2),
                       max(rec(f1, c1, f2, j - 1), rec(f1, j + 1, f2, c2))));
      }
  return (f2 - f1 + 1) * (c2 - c1 + 1);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> H >> W;
  for (int i = 0; i < H; i++) {
    for (int j = 0; j < W; j++) {
      cin >> R[i][j];
    }
  }
  int a = rec(0, 0, H - 1, W - 1);
  cout << a << "\n";
  return 0;
}
