Problem:
Score : 300 points

Problem Statement

You are given a grid of squares with H horizontal rows and W vertical columns, where each square is painted white or black.
HW characters from A_{11} to A_{HW} represent the colors of the squares.
A_{ij} is # if the square at the i-th row from the top and the j-th column from the left is black, and A_{ij} is . if that square is white.

We will repeatedly perform the following operation until all the squares are black:

Every white square that shares a side with a black square, becomes black.

Find the number of operations that will be performed.
The initial grid has at least one black square.

Constraints

1 \leq H,W \leq 1000

A_{ij} is # or ..

The given grid has at least one black square.

Input

Input is given from Standard Input in the following format:

H W
A_{11}A_{12}...A_{1W}
:
A_{H1}A_{H2}...A_{HW}

Output

Print the number of operations that will be performed.

Sample Input 1

3 3
...
.#.
...

Sample Output 1

2

After one operation, all but the corners of the grid will be black. After one more operation, all the squares will be black.

Sample Input 2

6 6
..#..#
......
#..#..
......
.#....
....#.

Sample Output 2

3

Solution in Python (3.4.3) (judge verdict: Time Limit Exceeded)

#https://atcoder.jp/contests/agc033/tasks/agc033_a

HW = [int(i) for i in input().split()]
A = []
for i in range(HW[0]):
    A.append(list(str(input())))

B = []
for i in range(len(A)):
    B.append([])
    for j in range(len(A[i])):
        B[i].append(A[i][j])

count = 0
count2 = 0
flag = False
tmp = ['.'] * len(A[0])
while True:
    for i in range(len(A)):
        if i > 0:
            tmp = A[i-1]
        for j in range(len(A[i])):
            if A[i][j] == '.':
                if flag == True:
                    B[i][j] = '#'
                    flag = False
                else:
                    if tmp[j] == '#':
                        B[i][j] = '#'
                    if i < len(A) - 1 and A[i+1][j] == '#':
                        B[i][j] = '#'
                    elif j < len(A[i]) - 1 and A[i][j+1] == '#':
                        B[i][j] = '#'
            else:
                count += 1
                flag = True
    
    for i in range(len(A)):
        for j in range(len(A[i])):
            A[i][j] = B[i][j]
    if count == HW[0] * HW[1]:
        break
    else:
        count = 0
        count2 += 1

print(count2)