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;
  vector<vector<int>> c(H, vector<int>(W + 1, 1));
  for (int i = 0; i < H; i++) {
    for (int j = 0; j < W; j++) {
      cin >> c[i][j];
    }
  }
  vector<vector<int>> dp(H + 1, vector<int>(W + 1));
  for (int i = 0; i < H; i++) {
    for (int j = 0; j < W + 1; j++) {
      if (c[i][j]) {
        dp[i + 1][j] = 0;
      } else {
        dp[i + 1][j] = dp[i][j] + 1;
      }
    }
  }
  int res = 0;
  for (int i = 0; i < H; i++) {
    vector<int> st(1, 0), p(1, -1);
    for (int j = 0; j <= W; j++) {
      while (st.back() > dp[i + 1][j]) {
        res = max(res, st.back() * (j - p.back()));
        st.pop_back();
        p.pop_back();
      }
      if (st.back() < dp[i + 1][j]) {
        st.push_back(dp[i + 1][j]);
        p.push_back(j);
      }
    }
  }
  cout << res << endl;
  return 0;
}
