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
   seira3 */

#include<stdio.h>

int arrayIn(int obj,int size, int arr[]){ //objがarr内にあるかどうかを判別する関数,あれば1なければ0を返す
    int count = 0;
    for(int i = 0; i < size; i++){
        if(arr[i]==obj){
            count++;
            break;
        }
    }
    return count;
}

int main(void){
    int row, line, count, countRow, countLine; //変数宣言
    scanf("%d %d",&row,&line); //rowとlineに入力
    count = 0; //初期化
    countRow = 0; //初期化
    countLine = 0; //初期化
    char square[row][line]; //元の迷路の配列を宣言
    int whiteRow[row], whiteLine[line]; //行,列全て'.'の番号を記憶する配列
    for(int i = 0; i < row; i++){ //squareに入力
        for(int j = 0; j < line; j++){
            scanf(" %c",&square[i][j]);
        }
    }
    for(int i = 0; i < row; i++){ //全て-1に初期化(0で初期化すると0番目が全て'.'なときに区別がつかなくなる)
        whiteRow[i] = -1;
    }
    for(int i = 0; i < line; i++){ //全て-1に初期化(0で初期化すると0番目が全て'.'なときに区別がつかなくなる)
        whiteLine[i] = -1;
    }
    for(int i = 0; i < row; i++){ //何行目に全て'.'なのかを探してwhiteRowに格納する
        count = 0;
        for(int j = 0; j < line; j++){
            if (square[i][j] == '.'){
                count++;
            }
        }
        if(count == line){
            whiteRow[countRow] = i;
            countRow++;
        }
    }
    for(int i = 0; i < line; i++){ //何列目に全て'.'なのかを探してwhiteLineに格納する
        count = 0;
        for(int j = 0; j < row; j++){
            if (square[j][i] == '.'){
                count++;
            }
        }
        if(count == row){
            whiteLine[countLine] = i;
            countLine++;
        }
    }
    for(int i = 0; i < row; i++){ //答えを出力する
        for(int j = 0; j < line; j++){
            if(arrayIn(i,row,whiteRow) == 0 && arrayIn(j,line,whiteLine) == 0){ //もし全て'.'の行,列に属していなかったら
                printf("%c",square[i][j]);
            }
        }
        if(arrayIn(i,row,whiteRow) == 0){ //行が全て'.'でなければ改行
            printf("\n");
        }
    }
    return 0;
}