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;
struct cww{cww(){ios::sync_with_stdio(false);cin.tie(0);}}star;
#define P(x) cout << (x) << "\n"
#define p(x) cout << (x)
#define all(c) (c).begin(), (c).end()
#define rall(c) (c).rbegin(), (c).rend()
#define vv(type, c, m, n, i) vector<vector<type>> c(m, vector<type>(n, i));
#define rep(i,a,n) for(int i=(a), i##_len=(n); i<i##_len; ++i)
#define rrep(i,a,n) for(int i=(a), i##_len=(n); i>i##_len; --i)
#define len(x) ((int)(x).size())
#define mp make_pair
#define eb emplace_back
#define fi first
#define se second
#define chmax(a,b) if(a<b)a=b;
#define chmin(a,b) if(b<a)a=b;
typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<double> vd;
typedef vector<long long> vll;
typedef vector<string> vs;
typedef vector<bool> vb;
typedef pair<int, int> pii;
vector<pii> st;
void getLargestRectangle() {
	int H, W; cin >> H >> W;
	vv(int, v, H, W, 0);
	int x;
	rep(i, 0, H) rep(j, 0, W) {
		cin >> x;
		v[i][j] = !x;
	}
	rep(i, 0, H - 1) rep(j, 0, W) if(v[i + 1][j] != 0) v[i + 1][j] += v[i][j];
	int ans = 0, pos;
	rep(i, 0, H) {
		st.clear();
		st.eb(0, 0);
		rep(j, 0, W) {
			if (st.back().fi == v[i][j]) continue;
			pos = j;
			while (st.back().fi > v[i][j]) {
				chmax(ans, st.back().fi * (j - st.back().se));
				pos = st.back().se; st.pop_back();
			}
			st.eb(v[i][j], pos);
		}
		while (!st.empty()) {
			chmax(ans, st.back().fi * (W - st.back().se));
			st.pop_back();
		}
	}
	P(ans);
}
int main() {
	getLargestRectangle();
	return 0;
}