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 N = 1405, INF = 0x3f3f3f3f;
int n, m;
int a[N][N];
int s[N], p;
inline int nextInt(void) {
  int res;
  scanf("%d", &res);
  return res;
}
inline int DP(int x) {
  int ans = 0;
  p = 0;
  s[++p] = a[x][1];
  for (int i = 2; i <= m + 1; i++)
    if (a[x][i] >= s[p]) {
      s[++p] = a[x][i];
    } else {
      int t = p;
      while (t && s[t] > a[x][i]) {
        ans = max(ans, s[t] * (p - t + 1));
        t--;
      }
      s[p = t + 1] = a[x][i];
    }
  return ans;
}
signed main(void) {
  memset(a, 0, sizeof(a));
  n = nextInt();
  m = nextInt();
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++)
      if (!nextInt()) a[i][j] = a[i][j - 1] + 1;
  int ans = 0;
  for (int i = 1; i <= n; i++) ans = max(ans, DP(i));
  printf("%d\n", ans);
}
