#include <bits/stdc++.h>
using namespace std;
const int INF = INT_MAX;
const long long INFL = LLONG_MAX;
const int output_precision = 15;
const long double pi = acos(-1);
const bool debug = true;
int main() {
  ios_base::sync_with_stdio(0);
  cout.precision(output_precision);
  cout << fixed;
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int(i) = 0, j1234 = n; (i) < j1234; (i)++) {
    for (int(j) = 0, j1234 = m; (j) < j1234; (j)++) {
      cout << (i % 2 == 0 ||
                       ((j == m - 1 && i % 4 == 1) || (j == 0 && i % 4 == 3))
                   ? "#"
                   : ".");
    }
    cout << '\n';
  }
}

### Prompt

Generate a CPP solution to the following problem:

Fox Ciel starts to learn programming. The first task is drawing a fox! However, that turns out to be too hard for a beginner, so she decides to draw a snake instead.

A snake is a pattern on a n by m table. Denote c-th cell of r-th row as (r, c). The tail of the snake is located at (1, 1), then it's body extends to (1, m), then goes down 2 rows to (3, m), then goes left to (3, 1) and so on.

Your task is to draw this snake for Fox Ciel: the empty cells should be represented as dot characters ('.') and the snake cells should be filled with number signs ('#').

Consider sample tests in order to understand the snake pattern.

Input

The only line contains two integers: n and m (3 ≤ n, m ≤ 50). 

n is an odd number.

Output

Output n lines. Each line should contain a string consisting of m characters. Do not output spaces.

Examples

Input

3 3


Output

###
..#
###


Input

3 4


Output

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


Input

5 3


Output

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


Input

9 9


Output

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int INF = INT_MAX;
const long long INFL = LLONG_MAX;
const int output_precision = 15;
const long double pi = acos(-1);
const bool debug = true;
int main() {
  ios_base::sync_with_stdio(0);
  cout.precision(output_precision);
  cout << fixed;
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int(i) = 0, j1234 = n; (i) < j1234; (i)++) {
    for (int(j) = 0, j1234 = m; (j) < j1234; (j)++) {
      cout << (i % 2 == 0 ||
                       ((j == m - 1 && i % 4 == 1) || (j == 0 && i % 4 == 3))
                   ? "#"
                   : ".");
    }
    cout << '\n';
  }
}

```