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<cstdio>  
#include<algorithm>
#include<stack>
using namespace std;
const int maxx = 1410;

int H, W;
int buffer[maxx][maxx];
int T[maxx][maxx];
struct Rectangle { int height; int pos; };

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

	for (int i = 0; i <= size; i++) {
		Rectangle rect;
		rect.height = height[i];
		rect.pos = i;
		// 当S为空
		if (S.empty())
			S.push(rect);
		// 栈顶的高度小于当前的高度
		else if (S.top().height < rect.height)
			S.push(rect);
		// 栈顶的高度大于当前的高度
		else if (S.top().height > rect.height) {
			int target = i;
			// 栈顶的高度大于等于当前的高度
			// 出现小于栈顶高度的代表pre.pos~i之间都存在高度为pre.height的干净瓷砖
			// 此时可以计算当前的面积
			while (!S.empty() && S.top().height >= rect.height) {
				Rectangle pre = S.top(); S.pop();
				// 长 * 宽
				int area = pre.height * (i - pre.pos);
				maxv = max(maxv, area);
				target = pre.pos;
			}
			// 加上前面的宽度（下标设置为之前的），以便后面有合适的瓷砖进行计算
			rect.pos = target;
			S.push(rect);
		}
	}
	return maxv;
}

int solve() {
	// 以行为直方图,计算纵向的干净长度
	for (int j = 0; j < W; j++) {
		for (int i = 0; i < H; i++) {
			if (buffer[i][j])
				T[i][j] = 0;
			else
				T[i][j] = (i > 0) ? T[i - 1][j] + 1 : 1;
		}
	}

	// 从行计算出当前行的最大面积
	int maxv = 0;
	for (int i = 0; i < H; i++)
		maxv = max(maxv, getLargestRectangle(W, T[i]));
	return maxv;
}
int main()
{
	scanf("%d %d", &H, &W);
	for (int i = 0; i < H; i++)
		for (int j = 0; j < W; j++)
			scanf("%d", &buffer[i][j]);
	
	printf("%d\n", solve());
	return 0;
}
