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;
  char c;
  int mp[1010][1010] = {0};
  int ans = 0;
  cin >> h >> w;
  for (int i = 1; i <= h; i++) {
    for (int j = 1; j <= w; j++) {
      scanf(" %c", &c);
      if (c == '1') mp[i][j] = 0;
      if (c == '0') mp[i][j] = mp[i - 1][j] + 1;
    }
  }
  int mx;
  int target;
  stack<pair<int, int> > st;
  for (int i = 1; i <= h; i++) {
    mx = 0;
    for (int j = 1; j <= w; j++) {
      if (mx < mp[i][j]) {
        st.push(make_pair(mp[i][j], j));
        mx = mp[i][j];
      } else if (mx > mp[i][j]) {
        target = j;
        while (!st.empty()) {
          pair<int, int> f = st.top();
          if (f.first < mp[i][j]) break;
          st.pop();
          ans = max(ans, (f.first * (j - f.second)));
          target = f.second;
        }
        if (mp[i][j] > 0) st.push(make_pair(mp[i][j], target));
        mx = mp[i][j];
      }
    }
    while (!st.empty()) {
      pair<int, int> f = st.top();
      st.pop();
      ans = max(ans, (f.first * (w + 1 - f.second)));
    }
  }
  cout << ans << endl;
  return (0);
}
