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;
int dp[1405][1405];
int main(){
    int h, w; cin >> h >> w;
    for (int i = 0; i < h; i++)
    {
        for (int j = 0; j < w; j++)
        {
            bool c; cin >> c;
            dp[i][j] = !c;
        }
    }
    for (int i = 1; i < h; i++)
    {
        for (int j = 0; j < w; j++)
        {
            dp[i][j] = dp[i-1][j] * dp[i][j] + dp[i][j];
        }
    }
    
    stack<pair<int, int>> S;
    int ans = 0;
    for (int i = 0; i < h; i++)
    {
        for (int j = 0; j <= w; j++)
        {  
            if (S.empty()){S.push(make_pair(dp[i][j], j));continue;}

            if (S.top().first < dp[i][j])
            {
                S.push(make_pair(dp[i][j], j));
            }else if(S.top().first > dp[i][j])
            {
                int target = j;
                while ((!S.empty()) && S.top().first >= dp[i][j])
                {
                    pair<int,int> num = S.top();S.pop();
                    ans = max(ans, (j-num.second) * num.first);
                    target = num.second;
                }
                S.push(make_pair(dp[i][j], target));
            }
        }
        
    }
    cout << ans << endl;
    
}
