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++14 (GCC 5.4.1) (judge verdict: Accepted)

// #define _CRT_SECURE_NO_WARNINGS
// #define _USE_MATH_DEFINES	// M_PI=3.1415...
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <cstdlib>
#include <functional>
#include <locale>
#include <cctype>
#include <sstream>
#include <iomanip>	// 20桁出力 cout << setprecision(20) << double;
#include <queue>
#include <bitset>	// bitset<8> bs1(131uL); // 10000011 bs1[0]は1 01stringからビット集合生成可
#include <numeric>
#include <random>
#include <climits>  // A/Bを計算するときA==LLONG_MIN && B==-1のみ(1ull<<63)で代用
using namespace std;

typedef long long LL;
typedef pair<LL, LL> P;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<LL> VL;
typedef vector<VL> VVL;				// VVL v(100,VL(100,0));
typedef vector<pair<LL, LL> > VP;
typedef vector<string> VS;
typedef map<int, int> MAPII;
typedef multimap<int, char, greater<int> > MuMAPIC;
typedef multimap<int, string, greater<int> > MuMIS;
typedef pair<LL, pair<P, P> > PP;

#define INF 999999999999999997		// 少し加算したらオーバーフローする
#define MP make_pair
#define FAST_IO  cin.tie(0); ios::sync_with_stdio(false);
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define FORL(i,a,b) for(LL i=(a);i<(b);i++)
#define FOR_REV(i,a,b) for(int i=(a);i>=(b);i--)
#define foreach_it(type,it,c) for(type::iterator it=c.begin(),c_end=c.end();it!=c_end;++it)
#define FOR_ITR(d) for(auto itr=d.begin(),d_end=d.end();itr!=d_end;++itr)	// C++11
#define O(s) cout<<s<<endl;

#define DUMP_VVI(b) if(dbgF){FOR(i,0,b.size()){FOR(j,0,b[i].size())printf("%d ",b[i][j]);puts("");}}
#define D_OUT(str,value) if(dbgF){cout<<str<<" : "<<value<<endl;}

template<class T>T IN() { T d; cin >> d; return d; }

LL gcd(LL a, LL b) { return (b > 0) ? gcd(b, a % b) : a; } // 最大公約数（Greatest Common Divisor）
LL lcm(LL a, LL b) { return a / gcd(a, b) * b; } // 最小公倍数（Least Common Multiple）
bool uruu(LL Y) { return (((Y % 4 == 0 && Y % 100 != 0) || Y % 400 == 0) ? true : false); } // うるう年判定
string Replace(string S, string from, string to) { auto p = S.find(from); while (p != string::npos) { S.replace(p, from.size(), to); p = S.find(from, p + to.size()); }return S; }
// 桁和
LL Ketawa_s(string s) { LL a = 0; FOR(i, 0, s.length())a += s[i] - '0'; return a; }

//二項係数 nCk mod pを求める
const LL MAX = 510000, p = 998244353; LL fac[MAX], finv[MAX], inv[MAX];
/* 前処理 */void COM_init() { fac[0] = fac[1] = 1; finv[0] = finv[1] = 1; inv[1] = 1; FOR(i, 2, MAX) { fac[i] = fac[i - 1] * i % p; inv[i] = p - inv[p % i] * (p / i) % p; finv[i] = finv[i - 1] * inv[i] % p; } }
/* 主処理 */LL COM(int n, int k) { if (n < k) return 0; if (n < 0 || k < 0) return 0; return fac[n] * (finv[k] * finv[n - k] % p) % p; }

int dx[4] = { 0,1,0,-1 };
int dy[4] = { 1,0,-1,0 };
int dxx[8] = { 0,1,1,1,0,-1,-1,-1 };
int dyy[8] = { 1,1,0,-1,-1,-1,0,1 };

// -------------------------------------------------------------------------------

int main() {
	FAST_IO;
	bool dbgF = true; // for D_OUT(str, value)  ... cout<< str <<" : "<< value <<endl;

	LL H, W;
	cin >> H >> W;

	VS map;
	VS ans;

	FOR(i, 0, H) {
		map.push_back(IN<string>());
	}

	FOR(i, 0, H) {
		bool flg = false;
		FOR(j, 0, W) {
			if (map[i][j] == '#') {
				flg = true;
				break;
			}
		}
		if (flg) {
			ans.push_back(map[i]);
		}
	}

	string initS;
	FOR(i, 0, W)
		initS += " ";

	VS ans2(ans.size(), initS);
	LL nextPushColomn = 0;

	FOR(i, 0, ans2[0].length()) {
		bool flg = false;
		FOR(j, 0, ans2.size()) {
			if (ans[j][i] == '#') {
				flg = true;
				break;
			}
		}

		if (flg) {
			FOR(j, 0, ans2.size()) {
				ans2[j][nextPushColomn] = ans[j][i];
			}
			nextPushColomn++;
		}
	}

	FOR(i, 0, ans2.size()) {
		FOR(j, 0, nextPushColomn) {
			cout << ans2[i][j];
		}
		cout << endl;
	}

	return 0;
}