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 main() {
  int h, w;
  cin >> h >> w;
  vector<vector<int>> v(h, vector<int>(w));
  for (long long i = 0; i < (long long)(h); i++)
    for (long long j = 0; j < (long long)(w); j++) cin >> v[i][j];
  for (long long i = 0; i < (long long)(h); i++)
    for (long long j = 0; j < (long long)(w); j++) v[i][j] = !v[i][j];
  static int dp[1500][1500];
  for (long long i = 0; i < (long long)(1500); i++)
    for (long long j = 0; j < (long long)(1500); j++) dp[i][j] = 0;
  for (long long i = 0; i < (long long)(w); i++) dp[0][i] = v[0][i];
  for (int i = 1; i < h; i++) {
    for (long long j = 0; j < (long long)(w); j++) {
      if (v[i][j] != 0) dp[i][j] = dp[i - 1][j] + v[i][j];
    }
  }
  int ans = 0;
  for (long long i = 0; i < (long long)(h); i++) {
    stack<pair<int, int>> st;
    for (long long j = 0; j < (long long)(w); j++) {
      pair<int, int> tmp = pair<int, int>(dp[i][j], j);
      if (st.empty())
        st.push(tmp);
      else if (st.top().first < tmp.first) {
        st.push(tmp);
      } else if (st.top().first >= tmp.first) {
        int used;
        while (st.size() && st.top().first >= tmp.first) {
          ans = max(ans, (tmp.second - st.top().second + 1) * tmp.first);
          used = st.top().second;
          st.pop();
        }
        tmp.second = used;
        st.push(tmp);
      }
    }
  }
  cout << ans << endl;
}
