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# include <bits/stdc++.h>
/*
测试用例：
5 6
1 0 1 0 0 0
0 1 0 1 0 0
1 0 0 0 1 0
0 1 0 0 0 1
1 0 1 0 0 0
答案
4
测试用例II：
5 6
1 0 1 0 0 0
0 1 0 1 1 0
1 0 0 0 1 0
0 1 0 0 0 1
1 0 1 0 0 0
答案
2
*/
using namespace std ;
const int N = 110 ;
int a [ N ] [ N ] ;
int s [ N ] [ N ] ;
int res ;
int main ( ) {
int n , m ;
cin > > n > > m ;
for ( int i = 1 ; i < = n ; i + + )
for ( int j = 1 ; j < = m ; j + + ) {
cin > > a [ i ] [ j ] ;
//维护二维前缀和
s [ i ] [ j ] = s [ i - 1 ] [ j ] + s [ i ] [ j - 1 ] - s [ i - 1 ] [ j - 1 ] + a [ i ] [ j ] ;
}
//直接在原数组上递推
for ( int i = n - 1 ; i > = 1 ; i - - ) //从倒数第二行开始
for ( int j = 1 ; j < = m ; j + + ) //这个需要考虑一下边界问题
if ( a [ i ] [ j ] ) { //只有当前位置大于0才有考虑的价值
a [ i ] [ j ] + = a [ i + 1 ] [ j + 1 ] ; //递推式,当前位置可以向下扩展的最长对角线长度
//利用二维前缀和计算区域的总和
int x1 = i , y1 = j , x2 = i + a [ i ] [ j ] - 1 , y2 = j + a [ i ] [ j ] - 1 ;
int cnt = s [ x2 ] [ y2 ] - s [ x1 - 1 ] [ y2 ] - s [ x2 ] [ y1 - 1 ] + s [ x1 - 1 ] [ y1 - 1 ] ;
if ( cnt = = a [ i ] [ j ] )
res = max ( res , a [ i ] [ j ] ) ; //在递推过程中不断求最大值
}
//输出
cout < < res < < endl ;
return 0 ;
}
