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 <iostream>
#include <fstream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <algorithm>
using namespace std;

#define REP(i,n) for(int i=0; i<n; ++i)
#define FOR(i,a,b) for(int i=a; i<=b; ++i)
#define FORR(i,a,b) for (int i=a; i>=b; --i)
#define ALL(c) (c).begin(), (c).end()

typedef long long ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<VI> VVI;
typedef vector<VL> VVL;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;

int max_rect(VI a){
    int n = a.size();
    VI l(n), r(n);

    stack<int> st;
    REP(i,n){
        while (!st.empty() && a[st.top()] >= a[i]) st.pop();
        l[i] = (st.empty() ? 0 : st.top() + 1);
        st.push(i);
    }

    while (!st.empty()) st.pop();
    FORR(i,n-1,0){
        while (!st.empty() && a[st.top()] >= a[i]) st.pop();
        r[i] = (st.empty() ? n : st.top());
        st.push(i);
    }

    int ret = 0;
    REP(i,n){
        ret = max(ret, a[i] * (r[i] - l[i]));
    }
    return ret;
}

int main(){
    int h, w;
    cin >> h >> w;

    VVI a(h, VI(w));
    REP(i,h) REP(j,w) cin >> a[i][j];

    VVI b(h, VI(w));
    REP(j,w){
        REP(i,h){
            if (a[i][j] == 0){
                b[i][j] = (i == 0 ? 1 : b[i-1][j] + 1);
            }
        }
    }

    int ans = 0;
    REP(i,h) ans = max(ans, max_rect(b[i]));
    cout << ans << endl;

    return 0;
}