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: Wrong Answer)

/* ex5_3
    riverstone*/

#include<stdio.h>
#include <stdbool.h>

int main(void){
    
    int height, width, i, j;
    
    //入力の受け取り
    scanf("%d %d", &height, &width);

    //多次元配列の定義
    //rowと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[height][width];
    bool row[height], col[width];

    //rowとcolをすべてfalseの配列にする
    for (int i = 0; i < height; i++) {
        row[i] = false;
    }
    for (int i = 0; i < height; i++) {
        col[i] = false;
    }

    //行列のマス(i,j)を読み取る
    //各行と各列に1つ以上の黒が存在するかをrowとcolに記録
    for(i = 0; i < height; i++){
        for(j = 0; j < width; j++){
            scanf(" %c", &block[i][j]);
            if(block[i][j] == '#'){
                //マス(i,j)が黒のとき,row[i],col[j]をtrueにする
                row[i] = true;
                col[j] = true;
            }
        }   
    }
    
    //rowとcolを見て共にtrueであるようなマスの場合のみ出力
    for(i = 0; i < height; i++){
        if(row[i] == 1){
            for(j = 0; j < width; j++){
                if(col[j] == 1){
                    //row[i] = tureかつcol[j] = falseのときblock[i][j]を表示
                    printf("%c", block[i][j]);
                }
            }
            //1行すべて処理が終わると改行する
            printf("\n");  
        } 
    }
 
    return 0;
}