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;
  char mat[h][w];
  int dp[h][w];
  for (int i = 0; i < h; i++) {
    for (int j = 0; j < w; j++) {
      cin >> mat[i][j];
    }
  }
  for (int i = 0; i < h; i++) dp[0][i] = (mat[0][i] == '0');
  for (int i = 1; i < h; i++) {
    for (int j = 0; j < w; j++) {
      if (mat[i][j] == '0')
        dp[i][j] = dp[i - 1][j] + 1;
      else
        dp[i][j] = 0;
    }
  }
  int ans = -1;
  for (int i = 0; i < h; i++) {
    stack<pair<int, int>> s;
    for (int j = 0; j < w; j++) {
      if (s.empty()) {
        s.push({j, dp[i][j]});
      } else if (dp[i][j] > s.top().second) {
        s.push({j, dp[i][j]});
      } else if (dp[i][j] == s.top().second) {
      } else if (dp[i][j] < s.top().second) {
        int left;
        while (!s.empty() && dp[i][j] < s.top().second) {
          pair<int, int> now = s.top();
          s.pop();
          ans = max(ans, (j - now.first) * (now.second));
          left = now.first;
        }
        s.push({left, dp[i][j]});
      }
    }
    while (!s.empty()) {
      pair<int, int> now = s.top();
      s.pop();
      ans = max(ans, (h - now.first) * (now.second));
    }
  }
  cout << ans << endl;
  return 0;
}
