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)
#define srep(i,s,t) for (int i = s; i < t; ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
using namespace std;
typedef long long int ll;
typedef pair<int,int> P;
#define yn {puts("Yes");}else{puts("No");}
#define MAX_N 200005
#define MAX 3005

struct Rectangle {int height; int pos;};

int getLargestRectangle(int size, int buffer[]){
    stack<Rectangle> S;
    int maxv = 0;
    buffer[size] = 0;

    for(int i = 0; i <= size; i++){
        Rectangle rect;
        rect.height = buffer[i];
        rect.pos = i;
        if(S.empty()){
            S.push(rect);
        }else{
            if(S.top().height < rect.height){
                S.push(rect);
            }else if(S.top().height > rect.height){
                int target = i;
                while(!S.empty() && S.top().height >= rect.height){
                    Rectangle pre = S.top(); S.pop();
                    int area = pre.height * (i - pre.pos);
                    maxv = max(maxv, area);
                    target = pre.pos;
                }
                rect.pos = target;
                S.push(rect);
            }
        }
    }
    return maxv;
}

int H, W;
int buffer[MAX][MAX];
int T[MAX][MAX];

int getLargestRectangle(){
    for(int j = 0; j < W; j++){
        for(int i = 0; i < H; i++){
            if(buffer[i][j]){
                T[i][j] = 0;
            }else{
                T[i][j] = (i > 0) ? T[i-1][j] + 1 : 1;
            }
        }
    }

    int maxv = 0;
    for(int i = 0; i < H; i++){
        maxv = max(maxv, getLargestRectangle(W, T[i]));
    }

    return maxv;
}


int main() {
    cin >> H >> W;
    rep(i,H)rep(j,W)cin >> buffer[i][j];

    int ans = getLargestRectangle();

    cout << ans << endl;
    return 0;
}
 
 

