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 rect {
  int pos, height;
};
int main() {
  int H, W;
  cin >> H >> W;
  int T[H][W];
  for (int i = 0; i < H; i++) {
    for (int j = 0; j < W; j++) {
      int c;
      cin >> c;
      if (i == 0) {
        T[i][j] = (c ? 0 : 1);
      } else {
        T[i][j] = (c ? 0 : T[i - 1][j] + 1);
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < W; i++) {
    stack<rect> stk;
    for (int j = 0; j < H; j++) {
      int pos, height;
      pos = j, height = T[i][j];
      if (stk.empty() || stk.top().height < height) {
        stk.push((rect){pos, height});
      } else if (stk.top().height > height) {
        while (!stk.empty() && stk.top().height >= height) {
          rect pre = stk.top();
          stk.pop();
          ans = max(ans, pre.height * (j - pre.pos));
          pos = pre.pos;
        }
        stk.push((rect){pos, height});
      }
    }
  }
  cout << ans << endl;
}
