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<cstdio>
#include<cstring>
#include<algorithm>
#include<stack>
using namespace std;
int h, w, map[1401][1401], upmost[1401][1401], ans;
struct rect{
    int h, pos;
    rect(){}
    rect(int h, int pos){
        this->h = h;
        this->pos = pos;
    }
};
stack<rect> dp;


int main(){
    //freopen("in.txt", "r", stdin);
    scanf("%d %d", &h, &w);
    memset(upmost, 0, sizeof(upmost));
    for(int i=0; i<h; i++){
        for(int j=0; j<w; j++){
            scanf("%d", &map[i][j]);
            upmost[i][j] = 1 - map[i][j];
        }
        upmost[i][w] = 0;
    }
    for(int i=1; i<h; i++)
    for(int j=0; j<w; j++)
        if(upmost[i][j]!=0)
            upmost[i][j] += upmost[i-1][j];

/*
    for(int i=0; i<h; i++){
        for(int j=0; j<w; j++)
            printf("%d ", upmost[i][j]);
        printf("\n");
    }
*/
    ans = 0; 
    for(int i=0; i<h; i++){
        while(!dp.empty()) dp.pop();
        for(int j=0; j<w+1; j++){
            if(dp.empty())
                dp.push(rect(upmost[i][j], j));
            else if(dp.top().h > upmost[i][j]){
                int left_most = j;
                while(!dp.empty() && dp.top().h > upmost[i][j]){
                    ans = max(ans, (j-dp.top().pos)*dp.top().h);
                    left_most = dp.top().pos;
                    dp.pop();
                }
                dp.push(rect(upmost[i][j], left_most));
            }else if(dp.top().h < upmost[i][j])
                dp.push(rect(upmost[i][j], j));
        }
    }    

    printf("%d\n", ans);
    return 0;
}
