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 Rectangle {
  int height;
  int pos;
};
int getLargestRectangle(int size, int buffer[]) {
  stack<Rectangle> S;
  int maxv = 0;
  buffer[size] = 0;
  for (int i = 0; i <= size; i++) {
    Rectangle rect;
    rect.height = buffer[i];
    rect.pos = i;
    if (S.empty()) {
      S.push(rect);
    } else {
      if (S.top().height < rect.height) {
        S.push(rect);
      } else if (S.top().height > rect.height) {
        int target = i;
        while (!S.empty() && S.top().height >= rect.height) {
          Rectangle pre = S.top();
          S.pop();
          int area = pre.height * (i - pre.pos);
          maxv = max(maxv, area);
          target = pre.pos;
        }
        rect.pos = target;
        S.push(rect);
      }
    }
  }
  return maxv;
}
int H, W;
int buffer[1400][1400];
int T[1400][1400];
int getLargestRectangle() {
  for (int j = 0; j < W; j++) {
    for (int i = 0; i < H; i++) {
      if (buffer[i][j]) {
        T[i][j] = 0;
      } else {
        T[i][j] = (i > 0) ? T[i - 1][j] + 1 : 1;
      }
    }
  }
  int maxv = 0;
  for (int i = 0; i < H; i++) {
    maxv = max(maxv, getLargestRectangle(W, T[i]));
  }
  return maxv;
}
int main(void) {
  int H, W;
  scanf("%d %d", &H, &W);
  for (int i = 0; i < H; i++) {
    for (int j = 0; j < W; j++) {
      scanf("%d", &buffer[i][j]);
    }
  }
  cout << getLargestRectangle() << endl;
  return 0;
}
