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, a[501][501], ls, s[501][501];
int main() {
  while (cin >> h >> w, h) {
    for (int i = 0; i < h; i++)
      for (int j = 0; j < w; j++) cin >> s[i][j];
    memset(a, 0, sizeof(a));
    for (int i = 0; i < h; i++)
      for (int j = 0; j < w; j++) {
        if (!i && !s[i][j]) a[i][j]++;
        if (i && !s[i][j]) a[i][j] = a[i - 1][j] + 1;
      }
    int ans = 0;
    for (int i = 0; i < h; i++) {
      stack<pair<int, int> > st;
      for (int j = 0; j < w + 1; j++) {
        if (st.empty()) st.push(pair<int, int>(a[i][j], j));
        if (st.top().first == a[i][j])
          continue;
        else if (st.top().first < a[i][j])
          st.push(pair<int, int>(a[i][j], j));
        else {
          while (!st.empty() && st.top().first > a[i][j]) {
            ans = max(ans, (j - st.top().second) * st.top().first);
            ls = st.top().second;
            st.pop();
          }
          st.push(pair<int, int>(a[i][j], ls));
        }
      }
    }
    cout << ans << endl;
  }
}
