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"
#define LF '\n'
#define ALL(x) x.begin(), x.end()
#define iostreamBooster() do{ cin.tie(nullptr); ios_base::sync_with_stdio(false); }while(0)
using namespace std;
typedef int64_t i64;
typedef pair<int,int> pii;
template<class A, class B>inline bool chmax(A &a, const B &b){return b>a ? a=b,1 : 0;}
template<class A, class B>inline bool chmin(A &a, const B &b){return b<a ? a=b,1 : 0;}
constexpr int INF = 0x3f3f3f3f;
template<class Itr> void dump(Itr begin, Itr end) { for (;begin != end; ++begin) clog << *begin << ' '; clog << LF;}

struct Rect { int height, left; Rect(int h=0, int l=0): height(h), left(l) {} };
int H, W;
int mat[1500][1500];
int dp[1500][1500];

int slv(const int *bar)
{
  stack<Rect> s;
  s.emplace(-1, -1);

  int largest = 0;
  for(int i = 0; i <= W; ++i) {
    const int h = bar[i];

    if(s.top().height < h) {
      s.emplace(h, i);
    }
    else if (s.top().height > h) {
      Rect r;
      while(s.top().height >= h) {
        r = s.top(); s.pop();
        chmax(largest, (i - r.left) * r.height);
      }
      s.emplace(h, r.left);
    }
  }

  return largest;
}

signed main()
{
  cin >> H >> W;
  for (int i = 0; i < H; ++i) {
    for(int j = 0; j < W; ++j) {
      cin >> mat[i][j];
      mat[i][j] = 1 - mat[i][j];
    }
  }

  int ans = 0;
  for (int i = H-1; i >= 0; --i) {
    for (int j = 0; j < W; ++j) {
      if (mat[i][j]) mat[i][j] += mat[i+1][j];
    }
    chmax(ans, slv(mat[i]));
  }

  cout << ans << endl;

  return 0;
}



