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<vector>

using namespace std;

int main()
{
	int H, W;
	cin >> H >> W;

	vector<vector<int>> map;
	for(int i = 0; i < H; i++)
	{
		vector<int> line;
		for(int j = 0; j < W  + 1; j++)
		{
			if(j == W)
			{
				line.push_back(0);
				continue;
			}
			int state;
			cin >> state;
			line.push_back(state == 0 ? 1 : 0);
		}
		map.push_back(line);
	}

	for(int i = 0; i < H-1; i++)
	{
		for(int j = 0; j < W; j++)
		{
			if(map[i + 1][j] != 0)
			{
				map[i + 1][j] += map[i][j];
			}
		}
	}
	int maxValue = 0;
	for(int i = 0; i < H; i++)
	{
		for(int j = 0; j < W; j++)
		{
			int minHight = map[i][j];
			if(minHight == 0)
			{
				continue;
			}
			for(int k = j + 1; k < W + 1; k++)
			{
				
				if(map[i][k] == 0)
				{
					int buf = (k-j) * minHight;
					maxValue = maxValue > buf ? maxValue : buf;
					break;
				}
				maxValue = maxValue > (k - j) * minHight ? maxValue : (k - j) * minHight;
				minHight = minHight < map[i][k] ? minHight : map[i][k];
				
			}
		}
	}
	cout << maxValue << endl;
}