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 <bits/stdc++.h>
using namespace std;

int main()
{
	int H, W;
	cin >> H >> W;
	vector<vector<int>> c(H, vector<int>(W + 1, 1));
	for (int i = 0; i < H; i++) {
		for (int j = 0; j < W; j++) {
			cin >> c[i][j];
		}
	}
	vector<vector<int>> dp(H + 1, vector<int>(W + 1));
	for (int i = 0; i < H; i++) {
		for (int j = 0; j < W + 1; j++) {
			if (c[i][j]) {
				dp[i + 1][j] = 0;
			}
			else {
				dp[i + 1][j] = dp[i][j] + 1;
			}
		}
	}
	int res = 0;
	for (int i = 0; i < H; i++) {
		vector<int> st(1, 0), p(1, -1);
		for (int j = 0; j <= W; j++) {
			int tp = j;
			while (st.back() > dp[i + 1][j]) {
				res = max(res, st.back() * (j - p.back()));
				tp = p.back();
				st.pop_back();
				p.pop_back();
			}
			if (st.back() < dp[i + 1][j]) {
				st.push_back(dp[i + 1][j]);
				p.push_back(tp);
			}
		}
	}
	cout << res << endl;
	return 0;
}

