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;
const int maxn = 1401;
int Squre[maxn][maxn];
struct rectangle {
  int height;
  int pos;
};
rectangle rec[maxn][maxn];
int solve(int h, int w) {
  int maxs = 0;
  for (int i = 0; i < h; i++) rec[i][w].height = 0;
  for (int i = 1; i < h; i++)
    for (int j = 1; j < w; j++) {
      if (Squre[i][j])
        rec[i][j].height = rec[i - 1][j].height + 1;
      else
        rec[i][j].height = 0;
      maxs = max(rec[i][j].height, maxs);
    }
  stack<rectangle> s;
  for (int i = 0; i < h; i++) {
    for (int j = 0; j <= w; j++) {
      rectangle rect;
      rect.height = rec[i][j].height;
      rect.pos = j;
      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 = j;
          while (!s.empty() && s.top().height >= rect.height) {
            rectangle pre = s.top();
            s.pop();
            int area = pre.height * (j - pre.pos);
            maxs = max(maxs, area);
            target = pre.pos;
          }
          rect.pos = target;
          s.push(rect);
        }
      }
    }
  }
  return maxs;
}
int main() {
  int h, w;
  cin >> h >> w;
  for (int i = 0; i < h; i++)
    for (int j = 0; j < w; j++) cin >> Squre[i][j];
  int ans = solve(h, w);
  cout << ans << endl;
  return 0;
}
