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;
size_t get_largest_rectangle(vector<vector<int>> &g) {
  size_t h = g.size(), w = g[0].size();
  vector<vector<size_t>> dp(h, vector<size_t>(w + 1));
  size_t max_area = 0;
  for (size_t i = 0; i < h; ++i)
    for (size_t j = 0; j < w; ++j) {
      dp[i][j] = !g[i][j];
      max_area |= dp[i][j];
    }
  for (size_t i = 1; i < h; ++i)
    for (size_t j = 0; j < w; ++j)
      if (dp[i][j]) dp[i][j] += dp[i - 1][j];
  for (size_t i = 0; i < h; ++i)
    for (size_t j = 0; j < w; ++j) {
      if (!dp[i][j]) continue;
      size_t min_height = dp[i][j];
      for (size_t k = j + 1; k <= w; ++k) {
        max_area = max(max_area, (k - j) * min_height);
        if (!dp[i][k]) break;
        if (min_height > dp[i][k]) min_height = dp[i][k];
      }
    }
  return max_area;
}
int main() {
  size_t h, w;
  scanf("%zu %zu", &h, &w);
  vector<vector<int>> g(h, vector<int>(w));
  for (size_t i = 0; i < h; ++i)
    for (size_t j = 0; j < w; ++j) scanf("%d", &g[i][j]);
  printf("%zu\n", get_largest_rectangle(g));
  return 0;
}
