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 G[2000][2000];
int T[2000][2000];
struct ang {
  int height, pos;
};
int f(int a[]) {
  stack<ang> S;
  int maxn = 0;
  for (int i = 0; i < W; i++) {
    ang rect;
    rect.height = a[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 pre = i;
        while (!S.empty() && S.top().height >= rect.height) {
          ang k = S.top();
          S.pop();
          int area = k.height * (i - k.pos);
          maxn = max(maxn, area);
          pre = k.pos;
        }
        rect.pos = pre;
      }
      S.push(rect);
    }
  }
  return maxn;
}
int main() {
  scanf("%d%d", &H, &W);
  for (int i = 0; i < H; i++) {
    for (int j = 0; j < W; j++) {
      scanf("%d", &G[i][j]);
    }
  }
  for (int j = 0; j < W; j++) {
    for (int i = 0; i < H; i++) {
      if (G[i][j])
        T[i][j] = 0;
      else
        T[i][j] = i > 0 ? T[i - 1][j] + 1 : 1;
    }
  }
  int maxn = 0;
  for (int i = 0; i < H; i++) {
    maxn = max(maxn, f(T[i]));
  }
  printf("%d\n", maxn);
  return 0;
}
