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++)
#define repr(i, n) for (int i = (int)(n); i >= 0; i--)
#define REP(i, m, n) for (int i = (int)(m); i <= (int)(n); i++)
#define REPR(i, m, n) for (int i = (int)(m); i >= (int)(n); i--)
#define all(v) v.begin(), v.end()
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
const int INF = 1001001001;

int H, W;
int field[1410][1410];

struct Rectangle { int height, pos; };

int main(){
  cin >> H >> W;

  rep(h, H){
    rep(w, W){
      char c; cin >> c;
      if(c == '0') field[h][w] = 1;
      else if(c == '1') field[h][w] = 0;
    }
  }
  rep(w, W){
    REP(h, 1, H-1){
      if(field[h][w] == 0) continue;
      field[h][w] += field[h-1][w];
    }
  }
  rep(h, H) field[h][W] = 0;

  int ma = 0;
  rep(h, H){
    stack<Rectangle> S;
    rep(w, W+1){
      Rectangle rect;
      rect.height = field[h][w];
      rect.pos = w;
      if(S.empty()) S.push(rect);
      else if(rect.height > S.top().height){
        S.push(rect);
      }
      else if(rect.height < S.top().height){
        int target = w;
        while(!S.empty() && rect.height <= S.top().height){
          Rectangle pre = S.top(); S.pop();
          int area = pre.height * (w - pre.pos);
          chmax(ma, area);
          target = pre.pos;
        }
        rect.pos = target;
        S.push(rect);
      }
    }
  }
 
  cout << ma << endl;

  return 0;

}


