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;
static const int MAX_N = 1400;
struct Rectangle {
  int height;
  int left;
};
int H, W;
int masu[MAX_N][MAX_N];
int dp[MAX_N][MAX_N];
int getLargestRectangle(int h) {
  stack<Rectangle> S;
  int maxArea = 0;
  for (int j = 0; j < W; j++) {
    Rectangle rect;
    rect.height = masu[h][j];
    rect.left = j;
    if (S.empty() || S.top().height < rect.height) {
      S.push(rect);
    } else if (S.top().height > rect.height) {
      int target = j;
      while (!S.empty() && S.top().height >= rect.height) {
        Rectangle pre = S.top();
        S.pop();
        int area = pre.height * (j - pre.left);
        maxArea = max(maxArea, area);
        target = pre.left;
      }
      rect.left = target;
      S.push(rect);
    }
  }
  return maxArea;
}
int main(void) {
  scanf("%d %d", &H, &W);
  for (int i = 0; i < H; i++) {
    for (int j = 0; j < W; j++) {
      scanf("%d", &masu[i][j]);
    }
  }
  for (int j = 0; j < W; j++) {
    masu[0][j] = (masu[0][j] + 1) % 2;
  }
  for (int i = 1; i < H; i++) {
    for (int j = 0; j < W; j++) {
      if (masu[i][j] == 0) {
        masu[i][j] = masu[i - 1][j] + 1;
      } else {
        masu[i][j] = 0;
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < H; i++) {
    ans = max(ans, getLargestRectangle(i));
  }
  printf("%d\n", ans);
  return 0;
}
