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>
#include <utility>

using namespace std;

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

	int maxArea = 0;
	for (int i = 1; i <= H; ++i) {
		stack<pair<int, int>> S;
		S.push(make_pair(0, V[i][0]));
		for (int j = 1; j <= W; ++j) {
			pair<int, int> p = S.top();
			if (V[i][j] > p.second) {
				S.push(make_pair(j, V[i][j]));
			} else if (V[i][j] < p.second) {
				int pos;
				while (!S.empty() && V[i][j] <= p.second) {
					S.pop();
					if (p.second > 0)
						pos = p.first;
					int area = (j - p.first) * p.second;
					maxArea = max(maxArea, area);
					if (!S.empty())
						p = S.top();
				}
				S.push(make_pair(pos, V[i][j]));
			}
		}
	}
	cout << maxArea << endl;

	return 0;
}