Problem:
Score : 700 points

Problem Statement

Snuke and Ciel went to a strange stationery store. Each of them got a transparent graph paper with H rows and W columns.

Snuke painted some of the cells red in his paper. Here, the cells painted red were 4-connected, that is, it was possible to traverse from any red cell to any other red cell, by moving to vertically or horizontally adjacent red cells only.

Ciel painted some of the cells blue in her paper. Here, the cells painted blue were 4-connected.

Afterwards, they precisely overlaid the two sheets in the same direction. Then, the intersection of the red cells and the blue cells appeared purple.

You are given a matrix of letters a_{ij} (1≤i≤H, 1≤j≤W) that describes the positions of the purple cells. If the cell at the i-th row and j-th column is purple, then a_{ij} is #, otherwise a_{ij} is .. Here, it is guaranteed that no outermost cell is purple. That is, if i=1, H or j = 1, W, then a_{ij} is ..

Find a pair of the set of the positions of the red cells and the blue cells that is consistent with the situation described. It can be shown that a solution always exists.

Constraints

3≤H,W≤500

a_{ij} is # or ..

If i=1,H or j=1,W, then a_{ij} is ..

At least one of a_{ij} is #.

Input

The input is given from Standard Input in the following format:

H W
a_{11}...a_{1W}
:
a_{H1}...a_{HW}

Output

Print a pair of the set of the positions of the red cells and the blue cells that is consistent with the situation, as follows:

The first H lines should describe the positions of the red cells.

The following 1 line should be empty.

The following H lines should describe the positions of the blue cells.

The description of the positions of the red or blue cells should follow the format of the description of the positions of the purple cells.

Sample Input 1

5 5
.....
.#.#.
.....
.#.#.
.....

Sample Output 1

.....
#####
#....
#####
.....

.###.
.#.#.
.#.#.
.#.#.
.....

One possible pair of the set of the positions of the red cells and the blue cells is as follows:

Sample Input 2

7 13
.............
.###.###.###.
.#.#.#...#...
.###.#...#...
.#.#.#.#.#...
.#.#.###.###.
.............

Sample Output 2

.............
.###########.
.###.###.###.
.###.###.###.
.###.###.###.
.###.###.###.
.............

.............
.###.###.###.
.#.#.#...#...
.###.#...#...
.#.#.#.#.#...
.#.#########.
.............

One possible pair of the set of the positions of the red cells and the blue cells is as follows:

Solution in Pascal (FPC 2.6.2) (judge verdict: Wrong Answer)

program ec12;
var 
	n,m,i,j,tot:longint;
	point:array[1..2510000,1..2] of longint;
	vis:array[1..501,1..501] of longint;
	a:array[1..501,1..501] of char;
procedure qsrt(l,r:longint);
var 
	i,j,v,s:longint;
begin 
	i:=l; j:=r; v:=point[(l+r) div 2][1];
	repeat
	while (point[i][1]<v) or ((point[i][1]=v) and (point[i][2]<point[(l+r) div 2][2])) do inc(i);
	while (point[j][1]>v) or ((point[j][1]=v) and (point[j][2]>point[(l+r) div 2][2])) do dec(j);
	if i<=j then 
	begin 
		s:=point[i][1];
		point[i][1]:=point[j][1];
		point[j][1]:=s;
		s:=point[i][2];
		point[i][2]:=point[j][2];
		point[j][2]:=s;
		inc(i);
		dec(j);
	end;
	until i>=j;
	if l<j then qsrt(l,j);
	if i<r then qsrt(i,r);
end;
begin 	
	fillchar(vis,sizeof(vis),0);
	readln(n,m);
	tot:=0;
	for i:=1 to n do
	begin 
		for j:=1 to m do
		begin 
			read(a[i,j]);	
			if a[i,j]='#' then 
			begin 
				inc(tot);
				vis[i,j]:=1;
				point[tot,1]:=i;
				point[tot,2]:=j;
			end;
		end;
		readln;
	end;
	qsrt(1,tot);
	for i:=1 to tot do 
	begin 
		if i>1 then 
		begin 
			if point[i][1]=point[i-1][1] then 
			begin 	
				for j:=point[i-1][2] to point[i][2] do 
				begin 
					if vis[point[i][1],j]=0 then 
					vis[point[i][1],j]:=2;
				end;
			end
			else
			begin 
				if point[i][2]=point[i-1][2] then 
				begin 
					for j:=point[i-1][1] to point[i][1] do 
					begin 	
						if vis[j,point[i][2]]=0 then
						vis[j,point[i][2]]:=2;
					end;
				end
				else
				begin 
					if point[i][2]<point[i-1][2] then 
					begin 
						for j:=point[i-1][1] to point[i][1] do 
						begin 
							if vis[j,point[i-1][2]]=0 then 
							vis[j,point[i-1][2]]:=2;
						end;
						for j:=point[i][2] to point[i-1][2] do
						begin 
							if vis[point[i][1],j]=0 then 
							vis[point[i][1],j]:=2;
						end;
					end
					else
					begin 
						for j:=point[i-1][1] to point[i][1] do 
						begin 
							if vis[j,point[i-1][2]]=0 then 
							vis[j,point[i-1][2]]:=2;
						end;
						for j:=point[i-1][2] to point[i][2] do 
						begin 
							if vis[point[i][1],j]=0 then
							vis[point[i][1],j]:=2;
						end;
					end;
				end;
			end;
		end;
	end;
	for i:=1 to n do 
	begin 
		for j:=1 to m do
		begin 
			if vis[i,j]>=1 then 
			write('#')
			else
			write('.');
		end;
		writeln;
	end;
	writeln;
	for i:=1 to n do 
	begin 	
		for j:=1 to m do
		begin 	
			if vis[i,j]<=1 then 
			write('#')
			else
			write('.');
		end;
		writeln;
	end;
end.