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;
#define rep(i,n) for(int i = 0; i < (int)n; i++)
using ll = long long;

int main(){
  int h, w;
  cin >> h >> w;
  int c[h][w];
  rep(i,h) rep(j,w) cin >> c[i][j];
  int hist[h][w];
  rep(i,w) hist[0][i] = (c[0][i] == 0 ? 1 : 0);
  for(int i = 1; i < h; i++)  rep(j,w)  hist[i][j] = (c[i][j] == 0 ?  hist[i-1][j] + 1 : 0);
  ll ans = 0;
  rep(x,h) {
    ll l[w], r[w], st[w];
    ll t = 0;
    rep(i,w) {
      while(t > 0 && hist[x][st[t-1]] >= hist[x][i]) t--;
      l[i] = (t == 0 ? 0 : st[t-1]+1);
      st[t++] = i;
    }
    t = 0;
    for(int i = w-1; i >= 0; i--) {
      while(t > 0 && hist[x][st[t-1]] >= hist[x][i]) t--;
      r[i] = (t == 0 ? w : st[t-1]);
      st[t++] = i; 
    }
    rep(i,w) ans = max(ans,hist[x][i]*(r[i]-l[i]));
  }
  cout << ans << endl;
  return 0;
}

