**Bargaining Table(maximum area rectangle in a square with 0 and 1 )

Bargaining Table

Bob wants to put a new bargaining table in his office. To do so he measured the office room thoroughly and drew its plan: Bob's office room is a rectangular room n × m meters. Each square meter of the room is either occupied by some furniture, or free. A bargaining table is rectangular, and should be placed so, that its sides are parallel to the office walls. Bob doesn't want to change or rearrange anything, that's why all the squares that will be occupied by the table should be initially free. Bob wants the new table to sit as many people as possible, thus its perimeter should be maximal. Help Bob find out the maximum possible perimeter of a bargaining table for his office.

Input

The first line contains 2 space-separated numbers n and m (1 ≤ n, m ≤ 25) — the office room dimensions. Then there follow n lines withm characters 0 or 1 each. 0 stands for a free square meter of the office room. 1 stands for an occupied square meter. It's guaranteed that at least one square meter in the room is free.

Output

Output one number — the maximum possible perimeter of a bargaining table for Bob's office room.

Sample test(s)

input

3 3
000
010
000

output

8

input

5 4
1100
0000
0000
0000
0000

output

16

----------------------editorial--------------------

In this problem one should find the maximal perimeter of a rectangle that contains no '1'. Define these rectangles "correct". To solve a problem you are to check each possible rectangle for correctness and calculate its perimeter. The easiest way to check all rectangles is using 6 nested cycles. Using 4 of them you fix the coordinates while other 2 will look for '1'. So the complexity is O((n*m)3). It seems slow, but those, who wrote such a solution, says that it hasn't any problems with TL.

One may interest in much faster solution. Using simple DP solution one can get a solution with an O((n*m)2) complexity. It's clear, that rectangle with coordinates (x1, y1, x2, y2) is correct if and only if rectangles (x1, y1, x2-1, y2) and (x1, y1, x2, y2-1) are correct, and board[x2][y2] = '0'. So each of rectangles can be checked in O(1) and totally there will be O((n*m)2) operations.

------------------------code-------------------------

#include <cstdio>
#include <algorithm>
using namespace std;

int main(){
    int n,m,s[26][26]={};
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++) for(int j=1;j<=m;j++){
        scanf("%1d",&s[i][j]);
        s[i][j]+=s[i-1][j]+s[i][j-1]-s[i-1][j-1];
    }
    int ans=0;
    for(int a=1;a<=n;a++) for(int b=1;b<=m;b++)
        for(int p=a;p<=n;p++) for(int q=b;q<=m;q++)
            if(s[p][q]-s[a-1][q]-s[p][b-1]+s[a-1][b-1]==0)
                ans=max(ans,(p+q-a-b+2)*2);
    printf("%d\n",ans);
}