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;
const int INF = 1 << 28;
int dp[1401][1401];
int solve(int H, int W) {
  for (int j = 0; j < (int)W; j++) {
    int v = 1;
    for (int i = 0; i < (int)H; i++) {
      if (!dp[i][j]) v = 0;
      dp[i][j] = v++;
    }
  }
  int ans = 0;
  for (int i = 0; i < (int)H; i++) {
    pair<int, int> p, p2;
    stack<pair<int, int> > s;
    for (int j = 0; j < (int)W; j++) {
      p = make_pair(dp[i][j], j);
      if (s.empty())
        s.push(p);
      else {
        if (s.top().first < p.first)
          s.push(p);
        else if (s.top().first > p.first) {
          int pnt = j;
          while (!s.empty() && s.top().first >= p.first) {
            p2 = s.top();
            s.pop();
            ans = max(ans, p2.first * (j - p2.second));
            pnt = p2.second;
          }
          p.second = pnt;
          s.push(p);
        }
      }
    }
  }
  return ans;
}
int main() {
  int H, W;
  cin >> H >> W;
  for (int i = 0; i < (int)H; i++)
    for (int j = 0; j < (int)W; j++) {
      cin >> dp[i][j];
      dp[i][j] = !dp[i][j];
    }
  cout << solve(H, W) << endl;
  return 0;
}
