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;
const int INF =
    sizeof(int) == sizeof(long long) ? 0x3f3f3f3f3f3f3f3fLL : 0x3f3f3f3f;
const int MOD = (int)(1e9) + 7;
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}
void largest_rectangle() {
  int H, W;
  cin >> H >> W;
  vector<vector<int>> v(H, vector<int>(W));
  for (int i = (0); i < (H); i++)
    for (int j = (0); j < (W); j++) {
      int x;
      cin >> x;
      v[i][j] = !x;
    }
  for (int i = (0); i < (H - 1); i++)
    for (int j = (0); j < (W); j++) {
      if (v[i + 1][j] == 0) continue;
      v[i + 1][j] += v[i][j];
    }
  using P = pair<int, int>;
  int ans = 0;
  for (int i = (0); i < (H); i++) {
    stack<P> st;
    st.emplace(0, 0);
    for (int j = (0); j < (W); j++) {
      ;
      if (st.top().first == v[i][j]) continue;
      while (st.top().first > v[i][j]) {
        chmax(ans, st.top().first * (j - st.top().second));
        st.pop();
      }
      st.emplace(v[i][j], j);
    }
    while (st.size()) {
      chmax(ans, st.top().first * (W - st.top().second));
      st.pop();
    }
  }
  cout << ans << endl;
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  largest_rectangle();
  return 0;
}
