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>> a(h, vector<int>(w));
  for (int i = 0; i < h; i++) {
    for (int j = 0; j < w; j++) {
      cin >> a[i][j];
      a[i][j] = !a[i][j];
    }
  }
  for (int i = 1; i < h; i++) {
    for (int j = 0; j < w; j++) {
      if (a[i][j] == 0) {
        continue;
      }
      a[i][j] += a[i - 1][j];
    }
    a[i].emplace_back(0);
  }
  int ans = 0;
  for (int i = 0; i < h; i++) {
    int tmp = 0;
    stack<pair<int, int>> st;
    for (int j = 0; j <= w; j++) {
      if (st.empty() || st.top().first < a[i][j]) {
        st.emplace(make_pair(a[i][j], j));
      } else {
        if (a[i][j] < st.top().first) {
          int target = j;
          while (!st.empty() && a[i][j] <= st.top().first) {
            int height = st.top().first;
            int pos = st.top().second;
            st.pop();
            int area = height * (j - pos);
            tmp = max(tmp, area);
            target = j;
          }
          st.emplace(make_pair(a[i][j], target));
        }
      }
    }
    ans = max(ans, tmp);
  }
  cout << ans << endl;
  return 0;
}
