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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
import java.awt.geom.Point2D;
import java.awt.*;
import static java.lang.System.*;
import static java.lang.Math.*;
//Loopで使う文字列の長さは固定！
//intで大丈夫？オーバーフローしない？
//Loop回数は本当に10⁹以内になってる？Loopは回る？？
public class Main {
    public static void main(String[]$){
        Scanner sc = new Scanner(in);
        int H=sc.nextInt();
        int W=sc.nextInt();
        int[][] a=new int[H][W];
        for (int i = 0; i < H; i++) {
            String s=sc.next();
            for (int j = 0; j < W; j++) {
                if(s.charAt(j)=='#')a[i][j]=1;
            }
        }
        int[] t=new int[W];
        for (int i = 0; i < H; i++) {
            int c=0;
            for (int j = 0; j < W; j++) {
                t[j]+=a[i][j];
                c+=a[i][j];
            }
            if(c==0){
                for (int j = 0; j < W; j++) {
                    a[i][j]=2;
                }
            }
        }
        for (int i = 0; i < W; i++) {
            if(t[i]==0){
                for (int j = 0; j < H; j++) {
                    a[j][i]=2;
                }
            }
        }
        for (int i = 0; i <H; i++) {
            int c=0;
            for (int j = 0; j < W; j++) {
                if(a[i][j]==0)out.print(".");
                if(a[i][j]==1)out.print("#");
                c+=a[i][j];
            }
            if(c!=2*W)out.println();
        }
    }
    static long power(long x,int n){
        long mod=1000000007;
        long ans=1;
        while (n>0) {
            if((n&1)==1){
                ans=(ans*x)%mod;
            }
            x=(x*x)%mod;
            n>>=1;
        }
        return ans;
    }
    //最大公約数
    static int gcd (int a, int b) {
        int temp;
        while((temp = a%b)!=0) {
            a = b;
            b = temp;
        }
        return b;
    }
//※最小公倍数はa*b/gcd(a,b)である
}