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 M[1500][1500];
int A[1500][1500];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) cin >> M[i][j];
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (!M[i][j]) {
        if (i == 0)
          A[i][j] = 1;
        else
          A[i][j] = A[i - 1][j] + 1;
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    stack<int> s;
    for (int j = 0; j < m; j++) {
      int cur = A[i][j];
      while (s.size() && A[i][s.top()] >= cur) s.pop();
      if (s.size() == 0) {
        ans = max(ans, cur * (j + 1));
      } else {
        ans = max(ans, cur * (j - s.top()));
      }
      s.push(j);
    }
    while (s.size()) s.pop();
    for (int j = m - 1; j >= 0; j--) {
      int cur = A[i][j];
      while (s.size() && A[i][s.top()] >= cur) s.pop();
      if (s.size() == 0) {
        ans = max(ans, cur * (m - j));
      } else {
        ans = max(ans, cur * (s.top() - j));
      }
      s.push(j);
    }
  }
  cout << ans << '\n';
  return 0;
}
