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
   KKOO  */

#include<stdio.h>

int main(void){
    int row, col, count;
    //マス目の行数、列数を入力
    scanf("%d %d", &row, &col);

    //マス目に入る文字を入力
    char grid[row][col];
    for(int i=0; i<row; i++){
        for(int j=0; j<col; j++){
            scanf(" %c", &grid[i][j]);
        }
    }

    //余分な行、列を記憶する配列
    int extra_row[row], extra_col[col];

    //配列の中身を全て1にする
    for(int i=0; i<row; i++){
        extra_row[i] = 1;
    }
    for(int j=0; j<col; j++){
        extra_col[j] = 1;
    }

    //ある行が全て'.'である場合はその行数を記憶
    for(int i=0; i<row; i++){
        for(int j=0; j<col; j++){
            if(grid[i][j] == '.'){
                count ++;
            }
        }
        if(count == col){
            //余分な行の要素は0にする
            extra_row[i] = 0;
        }
        count = 0;
    }

    //ある列が全て'.'である場合はその列数を記憶
    for(int j=0; j<col; j++){
        for(int i=0; i<row; i++){
            if(grid[i][j] == '.'){
                count ++;
            }
        }
        if(count == row){
            //余分な列の要素は0にする
            extra_col[j] = 0;
        }
        count = 0;
    }

    //配列の要素が1である部分のみ出力
    for(int i=0; i<row; i++){
        if(extra_row[i] == 1){
            for(int j=0; j<col; j++){
                if(extra_col[j] == 1){
                    printf("%c", grid[i][j]);
                }
            }
            printf("\n");
        }
    }

    return 0;
}