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<stack>
using namespace std;

int H, W, M[1500][1500], HIS[1500][1500];
struct rect { int h, pos; }r, pre;

int get(int his[])
{
	stack<rect>s;
	int M = 0;

	for (int i = 1; i <= W + 1; i++)
	{
		r.h = his[i], r.pos = i;

		if (s.empty())
			s.push(r);
		else
		{
			if (r.h >= s.top().h)
				s.push(r);
			else
			{
				int tar = i;

				while (!s.empty() && r.h <= s.top().h)
				{
					pre = s.top(), s.pop();
					tar = pre.pos;
					int area = pre.h * (i - pre.pos);
					M = max(M, area);
				}

				r.pos = tar;
				s.push(r);
			}
		}
	}

	return M;
}

int main()
{
	cin >> H >> W;
	
	for (int i = 1; i <= H; i++)
		for (int j = 1; j <= W; j++)
			cin >> M[i][j];
	
	for (int i = 1; i <= H; i++)
	{
		for (int j = 1; j <= W; j++)
		{
			if (M[i][j] == 0)
				HIS[i][j] = HIS[i - 1][j] + 1;
			else
				HIS[i][j] = 0;
		}
	}

	int MAX = 0;

	for (int i = 1; i <= H; i++)
		MAX = max(MAX, get(HIS[i]));

	cout << MAX << endl;
	return 0;
}
