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 melo */
#include <stdio.h>
#include <string.h>//文字列関数を読み取れるようにする
#include <stdlib.h>
int main(void){
    int row_N,col_N,i,j;//整数の箱を４つ用意する
    char arrayc[100][100];//100*100の文字配列を用意
    int row_TorF[100], col_TorF[100];//黒があるかないか判定する文字列を列と行分用意
    scanf("%d %d",&row_N,&col_N);//行と列を代入
    for(i=0;i<row_N;i++){//二次元の文字配列を代入
        for(j=0;j<col_N;j++){
            scanf(" %c",&arrayc[i][j]);
        }
    }
    for(i=0;i<row_N;i++){//初期化
        row_TorF[i]=0;
    }
    for(j=0;j<col_N;j++){//初期化
        col_TorF[j]=0;
    }
    for(i=0;i<row_N;i++){//黒がある行と列を探す
        for(j=0;j<col_N;j++){
            if(arrayc[i][j]=='#'){
                row_TorF[i]=1;//i列には黒がある
                col_TorF[j]=1;//i行には黒がある
            }
        }
    }
    for(i=0;i<row_N;i++){//白いマスのみからなる列または行以外（黒があった列または行）を表示する
        for(j=0;j<col_N;j++){
            if(row_TorF[i]==1&&col_TorF[j]==1){
                printf("%c",arrayc[i][j]);
            }
        }
        if(row_TorF[i]==1){
                printf("\n");//1列分表示されたら改行する
        }
    }
    return 0;//正常終了
}