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 <algorithm>
#include <cstdio>
#include <iostream>
#include <stack>

using namespace std;

struct Cell {
	int pos, height;
};

int main() {
	int H, W;
	scanf("%d %d", &H, &W);

	int A[H][W];
	for (int i = 0; i < H; ++i) {
		for (int j = 0; j < W; ++j) {
			scanf("%d", &A[i][j]);
		}
	}
	int C[H + 1][W + 1];
	for (int i = 0; i <= H; ++i) {
		C[i][W] = 0;
	}
	for (int i = 0; i <= W; ++i) {
		C[0][i] = 0;
	}

	for (int j = 0; j < W; ++j) {
		for (int i = 1; i <= H; ++i) {
			if (A[i - 1][j] == 0) {
				C[i][j] = C[i - 1][j] + 1;
			} else {
				C[i][j] = 0;
			}
		}
	}

	int maxArea = 0;
	for (int i = 1; i <= H; ++i) {
		stack<Cell> S;
		for (int j = 0; j <= W; ++j) {
			if (S.empty() || S.top().height < C[i][j]) {
				Cell c;
				c.height = C[i][j];
				c.pos = j;
				S.push(c);
			} else if (S.top().height > C[i][j]) {
				int prePos = j;
				while (!S.empty() && S.top().height > C[i][j]) {
					Cell pop = S.top();
					S.pop();
					int area = (j - pop.pos) * pop.height;
					maxArea = max(maxArea, area);
					prePos = pop.pos;
				}
				Cell c;
				c.height = C[i][j];
				c.pos = prePos;
				S.push(c);
			}
		}
	}
	cout << maxArea << endl;

	return 0;
}