from itertools import permutations
import sys

sys.setrecursionlimit(10 ** 6)
from bisect import *
from collections import *
from heapq import *

def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def SI(): return sys.stdin.readline()[:-1]
def LLI(rows_number): return [LI() for _ in range(rows_number)]
int1 = lambda x: int(x) - 1
def MI1(): return map(int1, sys.stdin.readline().split())
def LI1(): return list(map(int1, sys.stdin.readline().split()))
p2D = lambda x: print(*x, sep="\n")
dij = [(1, 0), (0, 1), (-1, 0), (0, -1)]

def main():
    h,w=MI()
    ss=[SI() for _ in range(h)]
    # 陸地に出来る最大正方形を見つける
    # 参考https://qiita.com/gushwell/items/4ab87b52f6511f11f58d
    # 左からそのマスまで連続する陸地の数を求める
    cc=[[0]*w for _ in range(h)]
    for i in range(h):
        for j in range(w):
            if ss[i][j]=="#":
                if j:cc[i][j]=cc[i][j-1]+1
                else:cc[i][j]=1
    # 上からそのマスまで連続する陸地の数を求める
    dd=[[0]*w for _ in range(h)]
    for j in range(w):
        for i in range(h):
            if ss[i][j]=="#":
                if i:dd[i][j]=dd[i-1][j]+1
                else:dd[i][j]=1
    # そのマスを右下とする正方形がとれる、最大の辺の長さ
    qq=[[0]*w for _ in range(h)]
    for i in range(h):
        for j in range(w):
            if cc[i][j]:
                if i*j:qq[i][j]=min(qq[i-1][j-1]+1,cc[i][j],dd[i][j])
                else:qq[i][j]=1
    #p2D(qq)
    mx=max(max(row) for row in qq)
    print((mx+1)//2)

main()