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
  kt125 */

#include<stdio.h>

int main(void){
    int height, width, //縦、横
         count; //'.'をカウントする変数
    scanf("%d %d", &height, &width);

    char board[height][width]; //与えられたマス目の配列
    for(int i=0; i < height;i++){
        for (int j=0; j < width; j++){
            scanf(" %c", &board[i][j]);
        }  
    }
  	int col_del[height], //消去する列の番号を記憶する配列
        row_del[width],  //消去する行の番号を記憶する配列
        k=0; //カウンタ変数

    for(int i=0; i < height; i++){
        for(int j=0; j < width; j++){
            if(board[i][j]== '.'){
                count++;
            }
        }
        if (count == width){ //i行が全て'.'だった場合
            row_del[k] = i; //配列の最初から順に用いるためにkを使う
            k++;
        }
        count = 0; //各行ごとにリセット
    }  
    int row_del_num = k; //消去する行の数

    k = 0; //リセット

    for(int i=0; i < width; i++){
        for(int j=0; j < height; j++){
            if(board[j][i]== '.'){
                count++;
            }
        }
        if (count == height){ //i列が全て'.'だった場合
            col_del[k] = i;
            k++;
        }
      	count = 0; //各列ごとにリセット
    } 

  	int col_del_num = k; //消去する列の数
    
    for(int i=0; i < row_del_num; i++){
        for (int j=0; j < width; j++){
            board[row_del[i]][j] = 'o'; //消去する行の文字を'o'に置き換える
        }
    }
    for(int i=0; i < col_del_num; i++){
        for (int j=0; j < height; j++){
            board[j][col_del[i]] = 'o'; //消去する列の文字を'o'に置き換える
        }
    }
  	k = 0; // リセット
  	
    char ans_row[height * width]; //答えとなるマス目の文字を一列に並べた配列
    for(int i=0; i < height; i++){
        for(int j=0; j < width; j++){
            if (board[i][j] != 'o'){ //'o'を飛ばす
                ans_row[k] = board[i][j];
              	k++;
            }
        }
    }

  	int height_ans = height - row_del_num, //答えとなる配列の縦の長さ
  		width_ans = width - col_del_num; //答えとなる配列の横の長さ

  	char ans[height_ans][width_ans];  //答えとなる配列

    for(int i=0; i < height_ans; i++){
        for(int j=0; j < width_ans;j++){
            ans[i][j] = ans_row[j + i * (width_ans)]; //一次元の配列をマス目に合うように二次元に直す
        }
    }
  
	for(int i=0; i < height_ans;i++){
        for (int j=0; j < width_ans; j++){
            printf("%c", ans[i][j]); //マス目をprint
        }
        printf("\n");
    }
    return 0;
}