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

........

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

........

#########

当遇到可以除二数时，要输出.,可以除二分两种情况，一为可整除4，一为不可，来判断点在左或在右。

#include<iostream>
using namespace std;
int main()
{
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
    {
        if (i % 2 == 0)
        {
            if (i / 2 % 2 == 0)
            {
                for (int j = 1; j <= m; j++)
                {
                    if (j == 1)
                    {
                        cout << '#';
                    }
                    else
                        cout << '.';
                    if (i == n && j == m)
                    {
                        break;
                    }
                    if (j == m)
                    {
                        cout << endl;
                    }
                }
            }
            else
                for (int j = 1; j <= m; j++)
                {
                    if (j == m)
                    {
                        cout << '#';
                    }
                    else
                        cout << '.';
                    if (i == n && j == m)
                    {
                        break;
                    }
                    if (j == m)
                    {
                        cout << endl;
                    }
                }
        }
        else
            for (int j = 1; j <= m; j++)
            {
                cout << '#';
                if (i == n && j == m)
                {
                    break;
                }
                if (j == m)
                    cout << endl;
            }
    }
    return 0;
}