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
   moka223711 */

#include <stdio.h>
#define ROW 101 // マス目の行の上限
#define COL 101 // マス目の列の上限

int main(void){
    // br = black_row, bc = black_column
    // 黒が少なくとも1マス含まれている行,列を記録する 0で初期化
    int row, col, i, j, br[ROW]={0}, bc[COL]={0};
    // マス目の配色を受け取る2次元配列
    char sq[ROW][COL];
    scanf("%d %d", &row, &col);
	
    for(i = 0; i < row; i++){
        for(j = 0; j < col; j++){
            // マス目の配色を受け取り、黒が1マスでも含まれている行と列はbr[i], bc[j]を1に変える
            scanf(" %c", &sq[i][j]);
            if(sq[i][j]=='#'){
                br[i] = 1;
                bc[j] = 1;
            }
        }
    }
    for(i = 0; i < row; i++){
        // 上から1行ずつ出力していく
        // 黒が1マスも含まれない行(すべて白の行)は飛ばす
        if(br[i] != 0){
            // このfor文が実行される行には黒が少なくとも1マス含まれている
            for(j = 0; j < col; j++){
                // 黒が1マスも含まれない列(すべて白の列)は飛ばす
                if(bc[j] != 0){
                    // マス目を出力
                    printf("%c", sq[i][j]);
                }
            }
            // 改行を出力
            printf("\n");
        }
    }
    return 0;
}