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>> a(h, vector<int>(w));
	for (int i = 0; i < h; i++) {
		for (int j = 0; j < w; j++) {
			cin >> a[i][j];
			a[i][j] = !a[i][j];
		}
	}
	for (int i = 1; i < h; i++) {
		for (int j = 0; j < w; j++) {
			if (a[i][j] == 0) {
				continue;
			}
			a[i][j] += a[i - 1][j];
		}
		a[i].emplace_back(0);
	}
	int ans = 0;
	for (int i = 0; i < h; i++) {
		int tmp = 0;
		stack<pair<int, int>> st;
		for (int j = 0; j <= w; j++) {
			if (st.empty() || st.top().first < a[i][j]) {
				st.emplace(make_pair(a[i][j], j));
			} else {
				if (a[i][j] < st.top().first) {
					int target = j;
					while (!st.empty() && a[i][j] <= st.top().first) {
						int height = st.top().first;
						int pos = st.top().second;
						st.pop();
						int area = height * (j - pos);
						tmp = max(tmp, area);
						target = pos;
					}
					st.emplace(make_pair(a[i][j], target));
				}
			}
		}
		ans = max(ans, tmp);
	}
	cout << ans << endl;
	return 0;
}