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
//By Vlgd
#include<bits/stdc++.h>
using namespace std;
const int N=1401;
int M[N][N],h,w;
struct rect{int h,pos;};
int maxrect(int r){
	stack<rect>S;
	int ans=0;
	for(int i=0;i<=w;i++){
		rect cur;
		cur.h=M[r][i],cur.pos=i;
		if(S.empty()) S.push(cur);
		else if(S.top().h<cur.h) S.push(cur);
		else if(S.top().h>cur.h){
			rect l;
			int left;
			while(!S.empty()&&S.top().h>=cur.h){
				l=S.top();S.pop();
				left=l.pos;
				int s=l.h*(cur.pos-l.pos);
				ans=max(ans,s);
			}
			cur.pos=left;
			S.push(cur);
		}
	}
	return ans;
}
void pre(){
	for(int i=0;i<=w;i++) M[0][i]=0;
	for(int i=1;i<=h;i++) M[i][w]=0;
	for(int i=1;i<=h;i++){
		for(int j=0;j<w;j++){
			if(M[i][j]) M[i][j]=0;
			else M[i][j]=M[i-1][j]+1;
		}
	}
}
int main(){
	int ans=0;
	scanf("%d%d",&h,&w);
	for(int i=1;i<=h;i++) for(int j=0;j<w;j++) scanf("%d",&M[i][j]);
	pre();
	for(int i=1;i<=h;i++) ans=max(ans,maxrect(i));
	printf("%d\n",ans);
//	for(int i=1;i<=h;i++){
//		for(int j=1;j<=w;j++) printf("%d ",M[i][j]);
//		printf("\n");
//	}
} 
