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 Cell {
  int pos, height;
};
int main() {
  int H, W;
  scanf("%d %d", &H, &W);
  int A[H + 1][W + 1];
  for (int i = 1; i <= H; ++i) {
    for (int j = 1; j <= W; ++j) {
      scanf("%d", &A[i][j]);
    }
  }
  int C[H + 1][W + 1];
  for (int i = 0; i <= H; ++i) {
    C[i][W] = 0;
  }
  for (int i = 0; i <= W; ++i) {
    C[0][i] = 0;
  }
  for (int j = 1; j <= W; ++j) {
    for (int i = 1; i <= H; ++i) {
      if (A[i][j] == 0) {
        C[i][j] = C[i - 1][j] + 1;
      } else {
        C[i][j] = 0;
      }
    }
  }
  int maxArea = 0;
  for (int i = 1; i <= H; ++i) {
    stack<Cell> S;
    for (int j = 1; j <= W; ++j) {
      Cell c;
      c.height = C[i][j];
      if (S.empty() || S.top().height < C[i][j]) {
        c.pos = j;
        S.push(c);
      } else if (S.top().height > C[i][j]) {
        int prePos = j;
        while (!S.empty() && S.top().height > C[i][j]) {
          Cell pop = S.top();
          S.pop();
          int area = (j - pop.pos) * pop.height;
          maxArea = max(maxArea, area);
          prePos = pop.pos;
        }
        c.pos = prePos;
        S.push(c);
      }
    }
  }
  cout << maxArea << endl;
  return 0;
}
