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

#define _USE_MATH_DEFINES
#define _CRT_SECURE_NO_WARNINGS
#include <bits/stdc++.h>
using namespace std;

//
#define int long long
#define pb(x) push_back(x)
#define m0(x) memset((x), 0, sizeof(x))
#define mm(x) memset((x), -1, sizeof(x))

//container
#define ALL(x) (x).begin(), (x).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define EACH(i, c) for (typeof((c).begin()) i = (c).begin(); i != (c).end(); ++i)
#define EXIST(s, e) ((s).find(e) != (s).end())
#define UNIQUE(v) (v).erase(unique((v).begin(), (v).end()), (v).end());
#define PERM(c) \
  sort(ALL(c)); \
  for (bool c##p = 1; c##p; c##p = next_permutation(ALL(c)))

// debug
#define GET_VAR_NAME(variable) #variable
#define test(x) cout << GET_VAR_NAME(x) << " = " << x << endl;

// bit_macro
#define bit(n) (1LL << (n))
#define bitset(a, b) (a) |= (1 << (b))
#define bitunset(a, b) (a) |= ~(1 << (b))
#define bitcheck(a, b) ((((a) >> (b)) & 1) == 1)
#define bitcount(a) __builtin_popcountll((a))

//typedef
typedef long long lint;
typedef complex<long double> Complex;
typedef pair<int, int> P;
typedef tuple<int, int, int> TP;
typedef vector<int> vec;
typedef vector<vec> mat;

//constant
const int INF = (int)1e18;
const int MOD = (int)1e9 + 7;
const double EPS = (double)1e-10;
const int dx[] = {-1, 0, 0, 1, 0, -1, -1, 1, 1};
const int dy[] = {0, -1, 1, 0, 0, 1, -1, 1, -1};

//
template <typename T>
void chmax(T &a, T b) { a = max(a, b); }
template <typename T>
void chmin(T &a, T b) { a = min(a, b); }
//
inline int toInt(string s) {
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T>
inline string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}

//
struct Accelerate_Cin {
  Accelerate_Cin() {
    cin.tie(0);
    ios::sync_with_stdio(0);
    cout << fixed << setprecision(20);
  };
};

signed main() {
  int H, W;
  cin >> H >> W;

  char red[510][510], blue[510][510], a[510][510];
  for (int i = 0; i < H; i++) {
    for (int j = 0; j < W; j++) {
      cin >> a[i][j];
      red[i][j] = '#';
      blue[i][j] = '.';
    }
  }

  for (int i = 1; i < H - 1; i += 2) {
    for (int j = 1; j < W - 1; j++) {
      blue[i][j] = '#';
      red[i][j] = '.';
    }
  }

  int cont = 0;
  for (int i = 2; i < H - 1; i += 2) {
    if (cont % 2 == 0) {
      blue[i][W - 2] = '#';
      red[i][W - 2] = '.';
    } else {
      blue[i][1] = '#';
      red[i][1] = '.';
    }
    cont++;
  }

  for (int i = 0; i < H; i++) {
    for (int j = 0; j < W; j++) {
      if (a[i][j] == '#') {
        blue[i][j] = red[i][j] = '#';
      }
    }
  }

  for (int i = 0; i < H; i++) {
    for (int j = 0; j < W; j++) {
      cout << red[i][j];
    }
    cout << endl;
  }

  cout << endl;

  for (int i = 0; i < H; i++) {
    for (int j = 0; j < W; j++) {
      cout << blue[i][j];
    }
    cout << endl;
  }
  return 0;
}