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 H, W;
int C[1410][1410];
int S[1410][1410];
deque<pair<int, int>> Q;
template <typename T>
T debug(const T& x) {
  cerr << x << " ";
  return x;
}
template <typename T>
T debugln(const T& x) {
  cerr << x << endl;
  return x;
}
int main() {
  iostream::sync_with_stdio(false);
  cin >> H >> W;
  for (int y = 0; y < H; y++)
    for (int x = 0; x < W; x++) cin >> C[y][x];
  for (int x = 0; x <= W; x++) S[0][x] = 0;
  for (int y = 0; y < H; y++) {
    for (int x = 0; x < W; x++)
      S[y + 1][x] = debug(C[y][x] == 0 ? S[y][x] + 1 : 0);
    cerr << endl;
  }
  int m = 0;
  for (int y = 0; y <= H; y++) {
    Q.emplace_front(-1, 0);
    for (int x = 0; x <= W; x++) {
      int last_pop = x;
      while (Q.front().second > S[y][x]) {
        m = max(m, (x - Q.front().first) * Q.front().second);
        last_pop = Q.front().first;
        Q.pop_front();
      }
      if (S[y][x] > Q.front().second) {
        Q.emplace_front(last_pop, S[y][x]);
        cerr << x << " " << S[y][x] << endl;
      }
    }
  }
  cout << m << endl;
  return 0;
}
