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<stdio.h>
#include<iostream>
#include<stack>
#include<algorithm>
#define MAX 1400

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[MAX][MAX];
int T[MAX][MAX];

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() {
  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;
}
