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;
#define MAX 1400

struct Rectangle {
  int height;
  int pos;
};

int getLargestRect(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);
    }
    if (S.top().height < rect.height) {
      S.push(rect);
    }
    else if (S.top().height == rect.height) {
      ;
    }
    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 getLargestRect() {
  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, getLargestRect(W, T[i]));
  }
  return maxv;
}

int main() {
  cin >> H >> W;
  for (int i = 0; i < H; i++) {
    for (int j = 0; j < W; j++) {
      cin >> buffer[i][j];
    }
  }

  cout << getLargestRect() << endl;

  return 0;
}

