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>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
typedef long long ll;

int T[1500][1500];
struct rect {
    int h;
    int p;
};

int main() {
    int H, W; cin >> H >> W;
    rep(i, H){
        rep(j, W){
            cin >> T[i][j];
            T[i][j] = (T[i][j]+1) % 2;
        }
    }

    for (int i = 1; i < H; i++){
        for (int j = 0; j < W; j++){
            if(T[i][j] != 0){
                T[i][j] = T[i-1][j]+1;
            }
        }
    }

    int ans = 0;
    rep(i, H){
        stack<rect> S;
        rep(j, W+1){
            rect r = {T[i][j], j};
            if(S.empty()){
                S.push(r);
            } else if(S.top().h < r.h){
                S.push(r);
            } else if(S.top().h == r.h){
                continue;
            } else {
                int nowp;
                while(!S.empty() && S.top().h > r.h){
                    rect r2 = S.top(); S.pop();
                    int men = r2.h * (j - r2.p);
                    ans = max(ans, men);
                    nowp = r2.p;
                }
                r.p = nowp;
                S.push(r);
            }
        }
    }

    cout << ans << endl;

    return 0;
}
