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 MAX = 1400;
int main() {
  int H, W;
  int tile[MAX][MAX];
  int ans = 0;
  scanf("%d%d", &H, &W);
  for (int i = 0; i < H; i++) {
    for (int j = 0; j < W; j++) {
      scanf("%d", &tile[i][j]);
    }
  }
  for (int y = 0; y < H; y++) {
    for (int x = W - 1; x >= 0; x--) {
      if (tile[y][x] == 1)
        tile[y][x] = -1;
      else if (x == W - 1)
        tile[y][x] = 0;
      else
        tile[y][x] = tile[y][x + 1] + 1;
    }
  }
  for (int x = 0; x < W; x++) {
    stack<pair<int, int>> s;
    s.push(pair<int, int>(-1, -1));
    for (int y = 0; y < H; y++) {
      if (s.top().second < tile[y][x]) {
        s.push(pair<int, int>(y, tile[y][x]));
      } else if (s.top().second > tile[y][x]) {
        while (s.top().second > tile[y][x]) {
          ans = max(ans, (y - s.top().first) * s.top().second);
          s.pop();
        }
        if (s.top().second != tile[y][x]) s.push(pair<int, int>(y, tile[y][x]));
      }
    }
    while (!s.empty()) {
      ans = max(ans, (W - s.top().first) * s.top().second);
      s.pop();
    }
  }
  printf("%d\n", ans);
  return 0;
}
