Problem:
Score : 200 points

Problem Statement

There is a grid of squares with H horizontal rows and W vertical columns.
The square at the i-th row from the top and the j-th column from the left is represented as (i, j).
Each square is black or white.
The color of the square is given as an H-by-W matrix (a_{i, j}).
If a_{i, j} is ., the square (i, j) is white; if a_{i, j} is #, the square (i, j) is black.

Snuke is compressing this grid.
He will do so by repeatedly performing the following operation while there is a row or column that consists only of white squares:

Operation: choose any one row or column that consists only of white squares, remove it and delete the space between the rows or columns.

It can be shown that the final state of the grid is uniquely determined regardless of what row or column is chosen in each operation.
Find the final state of the grid.

Constraints

1 \leq H, W \leq 100

a_{i, j} is . or #.

There is at least one black square in the whole grid.

Input

Input is given from Standard Input in the following format:

H W
a_{1, 1}...a_{1, W}
:
a_{H, 1}...a_{H, W}

Output

Print the final state of the grid in the same format as input (without the numbers of rows and columns); see the samples for clarity.

Sample Input 1

4 4
##.#
....
##.#
.#.#

Sample Output 1

###
###
.##

The second row and the third column in the original grid will be removed.

Sample Input 2

3 3
#..
.#.
..#

Sample Output 2

#..
.#.
..#

As there is no row or column that consists only of white squares, no operation will be performed.

Sample Input 3

4 5
.....
.....
..#..
.....

Sample Output 3

#

Sample Input 4

7 6
......
....#.
.#....
..#...
..#...
......
.#..#.

Sample Output 4

..#
#..
.#.
.#.
#.#

Solution in Lua (Lua 5.3.5) (judge verdict: Accepted)

local H, W = io.read("n", "n", "l")
local A = {}
for h=1,H do
  A[h] = {}
  for w=1,W do
    local a = io.read(1)
    A[h][w] = a
  end
  io.read("l")
end
local RH ={}
for h=1,H do
  local remove = true
  for w=1,W do
    if A[h][w] == '#' then
      remove = false
      break
    end
  end
  if remove then
    RH[h] = 1
  end
end
local RW = {}
for w=1,W do
  local remove = true
  for h=1,H do
    if A[h][w] == '#' then
      remove = false
      break
    end
  end
  if remove then
    RW[w] = 1
  end
end

for h=1,H do
  if not RH[h] then
    for w=1,W do
      if not RW[w] then
        io.write(A[h][w])
      end
    end
    print()
  end
end