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,x,y) for(int i=(x);i<(y);++i)
#define debug(x) #x << "=" << (x)
 
#ifdef DEBUG
#define _GLIBCXX_DEBUG
#define dump(x) std::cerr << debug(x) << " (L:" << __LINE__ << ")" << std::endl
#else
#define dump(x)
#endif

typedef long long int ll;
typedef pair<int,int> pii;
//template<typename T> using vec=std::vector<T>;

const int inf=1<<30;
const long long int infll=1LL<<58;
const double EPS=1e-9;
const int dx[]={1,0,-1,0},dy[]={0,1,0,-1};

template <typename T> ostream &operator<<(ostream &os, const vector<T> &vec){
	os << "[";
	for (const auto &v : vec) {
		os << v << ",";
	}
	os << "]";
	return os;
}

int get_largest_rectangle(int c[],int size){
    stack<pii> stk;
    int res=0;
    c[size]=0;
    
    rep(i,0,size+1){
        pii rect;
        rect.first=c[i];
        rect.second=i;
        if(stk.empty() or stk.top().first<rect.first) stk.push(rect);
        else if(stk.top().first>rect.first){
            int target=i;
            while(!stk.empty() and stk.top().first>=rect.first){
                pii pre=stk.top();
                stk.pop();
                int area=pre.first*(i-pre.second);
                res=max(res,area);
                target=pre.second;
            }
            rect.second=target;
            stk.push(rect);
        }
    }
    return res;
}

void solve(){
    int h,w;
    cin >> h >> w;
    
    static int c[1400][1401];
    rep(i,0,h) rep(j,0,w){
        cin >> c[i][j];
        c[i][j]=(c[i][j]+1)%2;
    }
    
    rep(i,1,h) rep(j,0,w) if(c[i-1][j] and c[i][j]) c[i][j]=c[i-1][j]+1;
    
    int ans=0;
    rep(i,0,h) ans=max(ans,get_largest_rectangle(c[i],w));
    cout << ans << endl;
}

int main(){
	std::ios::sync_with_stdio(false);
	std::cin.tie(0);
	solve();
	return 0;
}