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 <iostream>
#include <algorithm>
#include <queue>

using namespace std;

int H,W;
int C[1410][1410];
int S[1410][1410];

deque<pair<int,int>> Q; 
int main(){
	iostream::sync_with_stdio(false);
	cin>>H>>W;
	for(int y=0;y<H;y++)
		for(int x=0;x<W;x++)
			cin>>C[y][x];
	for(int x=0;x<=W;x++)S[0][x]=0;
	//for(int y=0;y<=H;y++)S[y][0]=0;
	for(int y=0;y<H;y++)
	{
		for(int x=0;x<W;x++)
			S[y+1][x]=C[y][x]==0?S[y][x]+1:0;
		//cerr<<endl;
	}
	int m=0;
	
	for(int y=0;y<=H;y++){
		Q.emplace_front(-1,0);
		for(int x=0;x<=W;x++){
			int last_pop=x;
			while(Q.front().second>S[y][x]){
				m=max(m,(x-Q.front().first)*Q.front().second);
				last_pop=Q.front().first;
				Q.pop_front();
			}
			if(S[y][x]>Q.front().second){
				Q.emplace_front(last_pop,S[y][x]);
				//cerr<<x<<" "<<S[y][x]<<endl;
			}
		}
	}
			
	cout<<m<<endl;
	return 0;
	
}