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>
#include <algorithm>
#include <stack>
#pragma warning( disable:4996 )
using namespace std;

#define GMAX 1400
int G[GMAX][GMAX];
int T[GMAX][GMAX];

struct Rectangle { int height; int pos; };

int getLargestRectangle(int size, int buffer[])
{
	int maxv = 0;
	stack<Rectangle> S;
	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 getLargestRectangle(int H, int W)
{
	for (int i = 0; i < H; i++) {
		for (int j = 0; j < W; j++) {
			if (G[i][j]) {
				T[i][j] = 0;
			}
			else {
				T[i][j] = (i > 0) ? T[i - 1][j] + 1 : 1;
			}
		}
	}
	int maxv = 0;
	for (int i = 0; i < H; i++) {
		maxv = max(maxv, getLargestRectangle(W, T[i]));
	}
	return maxv;
}

int main()
{
	int H, W;
	scanf("%d %d", &H, &W);
	for (int i = 0; i < H; i++) {
		for (int j = 0; j < W; j++) scanf("%d", &G[i][j]);

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