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 MAX 1400+1
#define BLOCK 1

struct Rectangle{ int height;  int pos; };

int H, W;

int getLargestRectangle( int size, int buffer[] )
{
	stack<Rectangle> S;
	int maxv = 0;
	buffer[size] = 0;

	for( int i = 0; i <= size; i++ ) {
		Rectangle rect;
		rect.height = buffer[i];
		rect.pos = i;
		if ( S.empty() ) {
			S.push( rect );
		} else {
			if ( S.top().height < rect.height ) {
				S.push( rect );
			} else if ( S.top().height > rect.height ) {
				int target = i;
				while ( !S.empty() && S.top().height >= rect.height ) {
					Rectangle pre = S.top(); S.pop();
					int area = pre.height * (i - pre.pos);
					maxv = max( maxv, area );
					target = pre.pos;
				}
				rect.pos = target;
				S.push(rect);
			}
		}
	}
	return maxv;
}

int buffer[MAX][MAX];
int T[MAX][MAX];

int getLargestRectangle()
{
	for( int i = 0; i < H; i++ ) {
		for( int j = 0; j < W; j++ ) {
			T[i][j] = 0;
		}
	}

	for( int j = 0; j < W; j++ ) {
		int sequence = 0;
		for( int i = 0; i < H; i++ ) {
			if( buffer[i][j] == BLOCK ) {
				sequence = T[i][j] = 0;
			}
			else {
				T[i][j] = ++sequence;
			}
		}
	}

	int maxv = 0;
	for( int i = 0; i < H; i++ ) {
		maxv = max( maxv, getLargestRectangle( W, T[i] ) );
	}

	return maxv;
}

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

	cout << getLargestRectangle() << endl;;

	return 0;
}

