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 C (GCC 5.4.1) (judge verdict: Accepted)

/* ex5_3
   masaaki52	*/
#include <stdio.h>
int main(void){
  int row,column,i,j,rblack[100]={0},cblack[100]={0};//rblack,cblackは各行・列の＃数を数える配列
  char matrix[100][100];
  scanf("%d %d",&row,&column);//行数、列数を入力
  for(i=0;i<row;i++){//matrix内全てに
    for(j=0;j<column;j++){
      scanf(" %c",&matrix[i][j]);//.#の入力
      if(matrix[i][j] == '#'){//matrix内のあるマスが#のとき
        rblack[i] += 1;//配列rblackの、＃のあった行番目に＋１
        cblack[j] += 1;//配列cblackの、＃のあった列番目に＋１
      }
    }
  }
  for(i=0;i<row;i++){//matrix内の全マス目について
    for(j=0;j<column;j++){
      if(rblack[i] != 0 && cblack[j] != 0){//そのマスのある行と列のどちらにも＃があれば
        printf("%c",matrix[i][j]);//マスの値を出力
      }
    }
    if(rblack != 0){//ある行が全て.でなければ
      printf("\n");//改行
    }
  }
  return 0;
}