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 M[1400][1400];
int dp[1400][1400] = {};
int main() {
  int ans = 0;
  int H, W;
  cin >> H >> W;
  for (int i = 0; i < (H); i++)
    for (int j = 0; j < (W); j++) cin >> M[i][j];
  for (int j = 0; j < (W); j++)
    if (M[0][j] == 0) dp[0][j] = 1;
  for (int i = (1); i < (H); i++)
    for (int j = 0; j < (W); j++)
      if (M[i][j] == 0) dp[i][j] = dp[i - 1][j] + 1;
  for (int i = 0; i < (H); i++) {
    stack<pair<int, int> > stk;
    for (int j = 0; j < (W); j++) {
      if (stk.empty() || stk.top().second < dp[i][j]) {
        stk.push(make_pair((j), (dp[i][j])));
      } else if (stk.top().second > dp[i][j]) {
        while (stk.top().second > dp[i][j]) {
          int rect = (j - stk.top().first) * stk.top().second;
          ans = max(ans, rect);
          stk.pop();
          if (stk.empty()) break;
        }
      }
    }
  }
  cout << ans << endl;
  return 0;
}
