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;
struct node {
  int x, y;
};
stack<node> d;
int a[1500][1500];
int main() {
  int n, m;
  while (cin >> n >> m) {
    int x;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        cin >> x;
        if (!x) a[i][j] += a[i - 1][j] + 1;
      }
    }
    node t;
    int maxs = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        t.x = a[i][j];
        t.y = j;
        if (d.size() == 0 || t.x > d.top().x)
          d.push(t);
        else {
          int l = d.top().y;
          while (d.size() && d.top().x >= t.x) {
            maxs = max(maxs, (l - d.top().y + 1) * d.top().x);
            d.pop();
          }
          d.push(t);
        }
      }
      int l = d.top().y;
      while (d.size() > 0) {
        maxs = max(maxs, (l - d.top().y + 1) * d.top().x);
        d.pop();
      }
    }
    cout << maxs << endl;
  }
}
