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 rectagle {
	int pos, height;
};
//13233
int getMax(int T[], int W) {

	stack<rectagle> S;
	int maxv = 0;
	for (int i = 0; i <= W; ++i) {
		rectagle r;
		r.pos = i;
		if (i != W) {
			r.height = T[i];
		} else {
			r.height = 0;
		}
		if (S.empty() || r.height > S.top().height) {
			S.push(r);
		} else {
			int prepos = i;
			while (!S.empty() && S.top().height > r.height) {
				rectagle pop = S.top();
				S.pop();
				int area = (i - pop.pos) * pop.height;
				prepos = pop.pos;
				maxv = max(maxv, area);
			}
			r.pos = prepos;
			S.push(r);
		}
	}

	return maxv;
}

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

	int maxv = 0;
	for (int i = 0; i < H; ++i) {
		maxv = max(maxv, getMax(T[i], W));
	}
	cout << maxv << endl;

	return 0;
}