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)

/*  5_3
    Y_Aizaki    */

#include <stdio.h>
#define LIM 100

int main(void){
    int row, col;
    // 行数、列数

    int matrix[LIM][LIM] = {0};
    // マス目入力を受け取る配列

    int midway[LIM][LIM] = {0};
    // マス目入力を行ごとに見て白いマスを消した配列

    int last[LIM][LIM] = {0};
    // マス目入力を列ごとに見て白いマスを消した配列

    int row_count = 0;
    // その行の'.'の数を数え上げる
    int col_count = 0;
    // その列の'.'と消した要素（0）を数え上げる

    char test;
    // 文字型の一時的な格納場所
    // '.'を1,'#'を2へと変換する

    scanf("%d%d", &row, &col);

    for (int i = 0; i < row; i++){
        for (int j = 0; j < col; j++){
            // matrixに'.','#'に応じた数字を入力するループ

            scanf(" %c", &test);
            if (test == '.'){
                matrix[i][j]=1;
            }
            else if (test == '#'){
                matrix[i][j]=2;
            }
        }
    }

    for (int i = 0; i < row; i++){
        // 列を消すループ
        // midwayを作成

        for (int j = 0; j < col; j++){
            if (matrix[i][j] == 1){
            row_count += 1;
            }
        }

        if (row_count != col){
            // 1のみで構成された列以外を移す
            for (int j = 0; j < col; j++){
                midway[i][j] = matrix[i][j];
            }
        }

        row_count = 0;
    }

    for (int j = 0; j < col; j++){
        // 行を消すループ
        // lastを作成

        for (int i = 0; i < row; i++){
            if (midway[i][j] == 1 || midway[i][j] == 0){
                col_count += 1;
            }
        }

        if (col_count != row){
            // 1のみで構成された行以外を移す
            for (int i = 0; i < row; i++){
                last[i][j] = midway[i][j];
            }
        }

        col_count = 0;
    }

    for (int i = 0; i < row; i++){
        for (int j = 0; j < col; j++){
            // lastを出力するループ

            if (last[i][j] == 1){
                printf(".");
            }
            else if (last[i][j] == 2){
                printf("#");
            }
        }

        for (int j = 0; j < col; j++){
            if (last[i][j] == 1 || last[i][j] == 2){
                printf("\n");
                break;
            }

        }
    }

    return 0;

}