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)

/* ex3-3
    kas056*/

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

int main(void){
    int i,j;
    int row, column;
    char **ptr;
    int *judge_r, *judge_c;
    scanf("%d %d" ,&row, &column);

    

    //ptrのメモリ確保
    ptr = (char**)calloc(row, sizeof(char*));

    if(ptr == NULL){
        exit(1);
    }
    
    for(i=0; i<row; i++){
        ptr[i] = (char*)calloc(column, sizeof(char));
        if(ptr[i] == NULL){
            exit(1);
        }
    }

    //judge_rのメモリ確保
    judge_r = (int*)calloc(row, sizeof(int));
    if(judge_r == NULL){
        exit(1);
    }

    //judge_cのメモリ確保
    judge_c = (int*)calloc(column, sizeof(int));

    if(judge_c == NULL){
        exit(1);
    }


    for(i=0; i<row; i++){
        for(j=0; j<column; j++){
            scanf(" %c", &ptr[i][j]);
        }
    }

    //＃があるか判定
    for(i = 0; i < row; i++){//まず行を固定して
        for (j = 0; j < column; j++){//その行の各要素に対して
            if(ptr[i][j] == '#'){//＃と一致するものがあったら終了する。
                break;
            }
        }

        if(j == column){//最後まで＃がなかったら
            judge_r[i] = 1;
        }
    }

    for(j = 0; j < column; j++){//列を固定して、行の場合と同様の操作を行う
        for(i = 0; i < row; i++){
            if(ptr[i][j] == '#'){
                break;
            }
        }
        if(i == row){
            judge_c[j] = 1;
        }
    }

    for(i=0; i<row; i++){//ある行において＃があった時、かつ
        if(judge_r[i] == 0){
            for(j = 0; j<column; j++){//その列に＃があったとき、
                if(judge_c[j] == 0){
                    printf("%c", ptr[i][j]);//(i,j)番目の文字を一文字出力
                }
            }//折り返し。これをcolumnの数だけ繰り返す
            printf("\n");//終点で改行
        }
    }//これをrowの数だけ繰り返す

    //逆順に開放
    for(i=0; i<row;i++){
        free(ptr[i]);
    }
    free(ptr);

    free(judge_r);
    free(judge_c);

    return 0;
}