#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
long long int add(long long int a, long long int b) {
  long long int ans = a + b;
  ans -= ans >= 1000000007 ? 1000000007 : 0;
  return ans;
}
long long int sub(long long int a, long long int b) {
  return (a - b + 1000000007) % 1000000007;
}
long long int mul(long long int a, long long int b) {
  return (a * b) % 1000000007;
}
long long int pow(long long int a, long long int b) {
  long long int ans = 1;
  while (b > 0)
    if (b & 1)
      ans = mul(ans, a), b--;
    else
      a = mul(a, a), b >>= 1;
  return ans;
}
long long int nCr(long long int n, long long int r) {
  if (r == 0) return 1;
  if (r == 1) return n;
  if (n == 0) return 0;
  return mul(n, mul(nCr(n - 1, r - 1), pow(r, 1000000007 - 2)));
}
const int Nmax = 1000005;
void solve() {
  int r, c;
  cin >> r >> c;
  int flag = 0;
  for (int i = 0; i < r; i++) {
    if (i % 2 == 0) {
      for (int i = 0; i < c; i++) {
        cout << "#";
      }
      cout << "\n";
    } else {
      if (flag == 0) {
        for (int i = 0; i < c - 1; i++) {
          cout << ".";
        }
        cout << "#"
             << "\n";
        flag = 1;
      } else {
        cout << "#";
        for (int i = 0; i < c - 1; i++) {
          cout << ".";
        }
        cout << "\n";
        flag = 0;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}

### Prompt

Create a solution in CPP for 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 double pi = acos(-1.0);
long long int add(long long int a, long long int b) {
  long long int ans = a + b;
  ans -= ans >= 1000000007 ? 1000000007 : 0;
  return ans;
}
long long int sub(long long int a, long long int b) {
  return (a - b + 1000000007) % 1000000007;
}
long long int mul(long long int a, long long int b) {
  return (a * b) % 1000000007;
}
long long int pow(long long int a, long long int b) {
  long long int ans = 1;
  while (b > 0)
    if (b & 1)
      ans = mul(ans, a), b--;
    else
      a = mul(a, a), b >>= 1;
  return ans;
}
long long int nCr(long long int n, long long int r) {
  if (r == 0) return 1;
  if (r == 1) return n;
  if (n == 0) return 0;
  return mul(n, mul(nCr(n - 1, r - 1), pow(r, 1000000007 - 2)));
}
const int Nmax = 1000005;
void solve() {
  int r, c;
  cin >> r >> c;
  int flag = 0;
  for (int i = 0; i < r; i++) {
    if (i % 2 == 0) {
      for (int i = 0; i < c; i++) {
        cout << "#";
      }
      cout << "\n";
    } else {
      if (flag == 0) {
        for (int i = 0; i < c - 1; i++) {
          cout << ".";
        }
        cout << "#"
             << "\n";
        flag = 1;
      } else {
        cout << "#";
        for (int i = 0; i < c - 1; i++) {
          cout << ".";
        }
        cout << "\n";
        flag = 0;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}

```