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;
  scanf("%d %d", &H, &W);
  int A[H][W];
  for (int i = 0; i < H; ++i) {
    for (int j = 0; j < W; ++j) {
      scanf("%d", &A[i][j]);
    }
  }
  int V[H + 1][W + 1];
  for (int i = 0; i <= H; ++i) {
    for (int j = 0; j <= W; ++j) {
      V[i][j] = 0;
    }
  }
  for (int j = 0; j < W; ++j) {
    for (int i = 1; i <= H; ++i) {
      if (A[i - 1][j] == 0)
        V[i][j] = V[i - 1][j] + 1;
      else {
        V[i][j] = 0;
      }
    }
  }
  int maxArea = 0;
  for (int i = 1; i <= H; ++i) {
    stack<pair<int, int>> S;
    S.push(make_pair(0, V[i][0]));
    for (int j = 1; j <= W; ++j) {
      pair<int, int> p = S.top();
      if (V[i][j] > p.second) {
        S.push(make_pair(j, V[i][j]));
      } else if (V[i][j] < p.second) {
        int pos;
        while (!S.empty() && V[i][j] <= p.second) {
          S.pop();
          if (p.second > 0) pos = p.first;
          int area = (j - p.first) * p.second;
          maxArea = max(maxArea, area);
          if (!S.empty()) p = S.top();
        }
        S.push(make_pair(pos, V[i][j]));
      }
    }
  }
  cout << maxArea << endl;
  return 0;
}
