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_2
   riverstone */

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

int main(void){
    
    int row, column, i, j;
    //多次元配列の定義
    //black_rowとblack_colは各行と各列に1つ以上の黒が存在するかを記録
    char **block;
    int *black_row, *black_col;
 
    //入力の受け取り
    scanf("%d %d", &row, &column);

    //列の分のメモリ領域
    block = (char**)calloc(row, sizeof(char*));
    if(block == NULL){
        exit(1);
    }

    // 各行のメモリ領域の確保
    for(i = 0; i < row; i++){
        block[i] = (char*)calloc(column, sizeof(char));
        if(block[i] == NULL){
            exit(1);
        }
    }

    black_row = (int*)calloc(row, sizeof(int));
    if(black_row == NULL){
            exit(1);
    }

    black_col = (int*)calloc(column, sizeof(int));
    if(black_col == NULL){
            exit(1);
    }

 
//行列の各マスを読み取る
    //各行と各列に1つ以上の黒が存在するかをblack_rowとblack_colに記録
    for(i = 0; i < row; i++){
        for(j = 0; j < column; j++){
            scanf(" %c", &block[i][j]);
            if(block[i][j] == '#'){
                //マス(i,j)が黒のとき,black_row[i],black_col[j]を1にする
                black_row[i] = 1;
                black_col[j] = 1;
            }
        }   
    }
    
    //black_rowとblack_colを見て共に1であるようなマスの場合のみ出力
    for(i = 0; i < row; i++){
        if(black_row[i] == 1){
            for(j = 0; j < column; j++){
                if(black_col[j] == 1){
                    //black_row[i] = 1かつblack_col[j] = 1のときblock[i][j]を表示
                    printf("%c", block[i][j]);
                }
            }
            printf("\n"); //1行の処理が全て終わると改行する
 
        } 
    }

    //各メモリの開放
    for(i=0; i<row; i++){
        free(block[i]);
    }
    free(block);
 
    free(black_col);
    free(black_row);

    return 0;
}