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;
const int N = 1402;
int n, m;
int G[N][N];
inline int nextInt(void) {
	int r; scanf("%d", &r); return r;
}
struct K {
	int h, c;
};
stack<K> st;
inline int maxRec(int d) {
	int ans = 0;
	for (int i = 1; i <= m; i++) {
		int h = G[d][i];
		int tmp = 0;
		while (!st.empty() && st.top().h >= h) {
			int newAns = st.top().h*(st.top().c + tmp);
			ans = max(ans, newAns);
			tmp += st.top().c;
			st.pop();
		}
		st.push((K) { h, 1 + tmp });
	}
	int tmp = 0;
	while (!st.empty()) {
		int newAns = st.top().h*(st.top().c + tmp);
		ans = max(ans, newAns);
		tmp += st.top().c;
		st.pop();
	}
	return ans;
}
signed main(void) {
	n = nextInt();
	m = nextInt();
	memset(G, 0, sizeof(G));
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++)
			if (!nextInt())
				G[i][j] = G[i-1][j] + 1;
	int ans = 0;
	for (int i = 1; i <= n; i++)
		ans = max(ans, maxRec(i));
	printf("%d\n", ans);
}