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;
long long int dy[] = {0, 0, 1, -1};
long long int dx[] = {1, -1, 0, 0};
long long int const MOD = 1000000007;
namespace std {
bool operator<(const complex<double>& a, const complex<double>& b) {
  return a.real() != b.real() ? a.real() < b.real() : a.imag() < b.imag();
}
}  // namespace std
long long int board[1500][1500];
long long int dp1[1500][1500];
long long int dp2[1500][1500];
signed main() {
  long long int H, W;
  cin >> H >> W;
  memset(dp1, 0, sizeof(dp1));
  memset(dp2, 0, sizeof(dp2));
  for (long long int(i) = (1); (i) <= (H); (i)++)
    for (long long int(j) = (1); (j) <= (W); (j)++) cin >> board[i][j];
  for (long long int(i) = (0); (i) <= (H); (i)++) board[i][0] = 1;
  for (long long int(i) = (0); (i) <= (W); (i)++) board[0][i] = 1;
  for (long long int(i) = (1); (i) <= (H); (i)++)
    for (long long int(j) = (1); (j) <= (W); (j)++) {
      if (board[i][j] == 1) {
        dp1[i][j] = 0;
        dp2[i][j] = 0;
      } else {
        dp1[i][j] = dp1[i - 1][j] + 1;
        dp2[i][j] = dp2[i][j - 1] + 1;
      }
    }
  long long int ans = 0;
  for (long long int(i) = (1); (i) <= (H); (i)++) {
    long long int t = 1, mi = 2000000000;
    for (long long int(j) = (1); (j) <= (W); (j)++) {
      long long int k = j - t + 1;
      if (dp1[i][j] == 0)
        t = j + 1;
      else {
        mi = min(mi, dp1[i][j]);
        ans = max(ans, mi * k);
      }
    }
  }
  for (long long int(j) = (1); (j) <= (W); (j)++) {
    long long int t = 1, mi = 2000000000;
    for (long long int(i) = (1); (i) <= (H); (i)++) {
      long long int k = i - t + 1;
      if (dp2[i][j] == 0)
        t = i + 1;
      else {
        mi = min(mi, dp2[i][j]);
        ans = max(ans, mi * k);
      }
    }
  }
  cout << ans << endl;
  return 0;
}
