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 9.2.1) (judge verdict: Accepted)

/* ex 7_3
   KitamoriFumiya */

#include <stdio.h>
#include <stdlib.h>

int main(void){
	int i, j, row, column, *ptr_row, *ptr_column;
	char **ptr;
	scanf("%d %d\n", &row, &column);
	//各行の配列へのポインタ分のメモリ領域を確保
	ptr = (char**)calloc(row, sizeof(char*));
	if(ptr == NULL){
		fprintf(stderr, "calloc error.\n");
		exit(1);
	}
	
	//各行の要素分のメモリ領域を確保
	for(i = 0; i < row; i++){
		ptr[i] = (char*)calloc(column, sizeof(char));
		if(ptr[i] == NULL){
			fprintf(stderr, "calloc error.\n");
			exit(1);
		}
	}
	// 残る行を記録するためのメモリ領域を確保
	ptr_row = (int*)calloc(row, sizeof(int));
	if(ptr_row == NULL){
		fprintf(stderr, "calloc error.\n");
		exit(1);
	}
	// 残る行を記録するptr_rowを初期化
	for(i = 0; i < row; i++){
		ptr_row[i] = 0;
	}
	// 残る列を記録するためのメモリ領域を確保
	ptr_column = (int*)calloc(column, sizeof(int));
	if(ptr_column == NULL){
		fprintf(stderr, "calloc error.\n");
	exit(1);
	}
	// 残る列を記録するptr_columnを初期化
	for(i = 0; i < column; i++){
		ptr_column[i] = 0;
	}
	// 元のマス目を入力し、黒を含む行と列は残すので残る列、行を記録する配列に1を格納
	for(i = 0; i < row; i++){
		for(j = 0; j < column; j++){
			scanf(" %c", &ptr[i][j]);
			if(ptr[i][j] == '#'){
				ptr_row[i] = 1;
				ptr_column[j] = 1;
			}
		}
	}
	// 黒のマスを含む行列を出力
	for(i = 0; i < row; i++){
		if(ptr_row[i] == 1){
			for(j = 0; j < column; j++){
				if(ptr_column[j] == 1){
					printf("%c", ptr[i][j]);
				}
			}
			printf("\n");
		}
	}
	
	// 残る行を記録するためのメモリ領域を開放
	free(ptr_row);
	// 残る列を記録するためのメモリ領域を開放
	free(ptr_column);
	
	//メモリ領域の解放，各行の要素分の領域を開放し，
	//各行の配列へのポインタの領域を解放
	//（メモリ領域の確保と逆順）
	for(i = 0; i < row; i++){
		free(ptr[i]);
	}
	free(ptr);
	
	return 0;
}