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;
struct elem {
  int pos, height;
};
int getArea(int V[], int W) {
  int area = 0;
  stack<elem> S;
  for (int i = 0; i <= W; ++i) {
    elem e;
    e.pos = i;
    if (i == W) {
      e.height = 0;
    } else {
      e.height = V[i];
    }
    if (S.empty() || e.height > S.top().height) {
      S.push(e);
    } else {
      int prepos = i;
      while (!S.empty() && S.top().height > e.height) {
        elem target = S.top();
        S.pop();
        area = max(area, (i - target.pos) * target.height);
      }
      e.pos = prepos;
      S.push(e);
    }
  }
  return area;
}
int main() {
  int H, W;
  scanf("%d %d", &H, &W);
  int A[H][W];
  for (int i = 0; i < H; ++i) {
    for (int j = 0; j < W; ++j) {
      scanf("%d", &A[i][j]);
    }
  }
  int V[H][W];
  for (int j = 0; j < W; ++j) {
    for (int i = 0; i < H; ++i) {
      if (A[i][j] == 1) {
        V[i][j] = 0;
      } else {
        if (i == 0) {
          V[i][j] = 1;
        } else {
          V[i][j] = V[i - 1][j] + 1;
        }
      }
    }
  }
  int maxv = 0;
  for (int i = 0; i < H; ++i) {
    maxv = max(maxv, getArea(V[i], W));
  }
  cout << maxv << endl;
  return 0;
}
