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 KitamoriFumiya */
#include<stdio.h>

// マス目を圧縮して、そのマス目を出力する関数
void compression(int height, int width, char matrix[height][width]){
	int row[height], col[width], i, j; // row[height](列に#が含まれているときに1を格納する配列),col[width](行に#が含まれているときに1を格納する配列),i,jをint型の変数として宣言
	
	// i行目に#が含まれていたらrowのi番目に1を記録
	for(i = 0; i < height; i++){
		row[i] = 0; // row[i]は初め0にしておく
		for(j = 0; j < width; j++){
			if(matrix[i][j] == '#'){
				row[i] = 1;
				break; // 記録したらループを抜け出す
			}
		}
	}
	
	// j行目に#が含まれていたらcolのj番目に1を記録
	for(j = 0; j < width; j++){
		col[j] = 0; // col[j]は初め0にしておく
		for(i = 0; i < height; i++){
			if(matrix[i][j] == '#'){
				col[j] = 1;
				break; // 記録したらループを抜け出す
			}
		}
	}
	
	// 1つでも#のある行と列を出力する
	for(i = 0; i < height; i++){
		// まず、i番目の行に#が含まれているかを判定
		if(row[i] == 1){
			for(j = 0; j < width; j++){
				// つぎに、j番目の列に#が含まれているかを判定し、含まれていたらmatrix[i]j[]を出力
				if(col[j] == 1){
					printf("%c", matrix[i][j]);
				}
			}
			printf("\n"); // 1行分を出力したら改行する
		}
	}
}

int main(void){
	int height, width, i, j; // height(行数),width(列数),i,jをint型の変数として宣言
	scanf("%d %d", &height, &width); // height,widthを入力
	char matrix[height][width]; // matrix[height][width](height行width列のマス目)を二次元配列の文字列として宣言
	// マス目の各要素を入力
	for (i = 0; i < height; i++){
		for (j = 0; j < width; j++){
			scanf(" %c", &matrix[i][j]);
		}
	}
	
	compression(height, width, matrix);
	
	return 0;
}