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)

/* ex7_3 
melo */
#include <stdio.h>
#include <stdlib.h>
int main(void){
    int row_N,col_N,i,j;//整数の箱を４つ用意する
    char **arrayc;//二重ポインタ
    int *row_TorF, *col_TorF;//黒があるかないか判別する
    scanf("%d %d",&row_N,&col_N);//行と列を代入
    arrayc=(char**)calloc(row_N,sizeof(char*));//列分のint*型ポイント配列を確保して確保した領域を0に初期化する
    if(arrayc==NULL){//確保できなかった場合エラーを出す
        fprintf(stderr,"memory allocation error.\n");
        exit(1);
    }
    for(int i=0;i<row_N;i++){
        arrayc[i]=(char*)calloc(col_N,sizeof(char));//行分のint型の配列を各ポイントに対して確保して確保した領域を0に初期化する
        if(arrayc[i]==NULL){//確保できなかった場合エラーを出す
            fprintf(stderr,"memory allocation error.\n");
            exit(1);
        }
    }
    row_TorF=(int*)calloc(row_N,sizeof(int));//領域を確保して確保した領域を0に初期化する
    if(row_TorF==NULL){//確保できなかった場合エラーを出す
        fprintf(stderr,"memory allocation error.\n");
        exit(1);
    }
    col_TorF=(int*)calloc(col_N,sizeof(int));//領域を確保して確保した領域を0に初期化する
    if(col_TorF==NULL){//確保できなかった場合エラーを出す
        fprintf(stderr,"memory allocation error.\n");
        exit(1);
    }
    for(i=0;i<row_N;i++){//二次元の文字配列を代入
        for(j=0;j<col_N;j++){
            scanf(" %c",&arrayc[i][j]);
        }
    }
    for(i=0;i<row_N;i++){//黒がある行と列を探す
        for(j=0;j<col_N;j++){
            if(arrayc[i][j]=='#'){
                row_TorF[i]=1;//i列には黒がある
                col_TorF[j]=1;//i行には黒がある
            }
        }
    }
    for(i=0;i<row_N;i++){//白いマスのみからなる列または行以外（黒があった列または行）を表示する
        for(j=0;j<col_N;j++){
            if(row_TorF[i]==1&&col_TorF[j]==1){
                printf("%c",arrayc[i][j]);
            }
        }
        if(row_TorF[i]==1){
                printf("\n");//1列分表示されたら改行する
        }
    }
    free(row_TorF);//メモリ解放
    free(col_TorF);//メモリ解放
    for(int i=0;i<row_N;i++){//二次元のメモリ解放
        free(arrayc[i]);
    }
    free(arrayc);
    return 0;//正常終了
}