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 <cstdio>           // printf(), getchar(), scanf()
#include <stack>

using namespace std;
static const int MAX_M = 1400;

struct Rectangle
{
	int height;
	int pos;
};

int G[MAX_M][MAX_M];
int dp[MAX_M][MAX_M + 1];
int H, W;

int
getLargestRectangle(int size, int H[])
{
	Rectangle rect;
	stack<Rectangle> S;

	int maxv = 0;
	H[size] = 0;

	for (int i = 0; i <= size; ++i)
	{
		rect.height = H[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);
					if (area > maxv)
						maxv = area;

					target = pre.pos;
				}

				rect.pos = target;
				S.push(rect);
			}
		}
	}

	return maxv;
}

int
getLargestRectangle()
{
	for (int j = 0; j < W; ++j)
	{
		for (int i = 0; i < H; ++i)
		{
			if (G[i][j])
				dp[i][j] = 0;
			else
				dp[i][j] = (i > 0) ? dp[i - 1][j] + 1 : 1;
		}
	}

	int maxv = 0;
	for (int i = 0; i < H; ++i)
	{
		int t = getLargestRectangle(W, dp[i]);
		if (t > maxv)
			maxv = t;
	}

	return maxv;
}

int
main(int argc, char** argv)
{
	int c;

	scanf("%d %d", &H, &W);
	c = getchar();
	for (int i = 0; i < H; ++i)
	{
		for (int j = 0; j < W; ++j)
		{
			while (c == ' ' || c == '\n')
				c = getchar();

			G[i][j] = (c != '0') ? 1 : 0;
			c = getchar();
		}
	}

	printf("%d\n", getLargestRectangle());
	return 0;
}