Problem:
Score : 500 points

Problem Statement

You are given two integers A and B.

Print a grid where each square is painted white or black that satisfies the following conditions, in the format specified in Output section:

Let the size of the grid be h \times w (h vertical, w horizontal). Both h and w are at most 100.

The set of the squares painted white is divided into exactly A connected components.

The set of the squares painted black is divided into exactly B connected components.

It can be proved that there always exist one or more solutions under the conditions specified in Constraints section.
If there are multiple solutions, any of them may be printed.

Notes

Two squares painted white, c_1 and c_2, are called connected when the square c_2 can be reached from the square c_1 passing only white squares by repeatedly moving up, down, left or right to an adjacent square.

A set of squares painted white, S, forms a connected component when the following conditions are met:

Any two squares in S are connected.

No pair of a square painted white that is not included in S and a square included in S is connected.

A connected component of squares painted black is defined similarly.

Constraints

1 \leq A \leq 500

1 \leq B \leq 500

Input

Input is given from Standard Input in the following format:

A B

Output

Output should be in the following format:

In the first line, print integers h and w representing the size of the grid you constructed, with a space in between.

Then, print h more lines. The i-th (1 \leq i \leq h) of these lines should contain a string s_i as follows:

If the square at the i-th row and j-th column (1 \leq j \leq w) in the grid is painted white, the j-th character in s_i should be ..

If the square at the i-th row and j-th column (1 \leq j \leq w) in the grid is painted black, the j-th character in s_i should be #.

Sample Input 1

2 3

Sample Output 1

3 3
##.
..#
#.#

This output corresponds to the grid below:

Sample Input 2

7 8

Sample Output 2

3 5
#.#.#
.#.#.
#.#.#

Sample Input 3

1 1

Sample Output 3

4 2
..
#.
##
##

Sample Input 4

3 14

Sample Output 4

8 18
..................
..................
....##.......####.
....#.#.....#.....
...#...#....#.....
..#.###.#...#.....
.#.......#..#.....
#.........#..####.

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

#include <bits/stdc++.h>
// #include <ext/pb_ds/assoc_container.hpp>/
// #include <ext/pb_ds/tree_policy.hpp>

#define sf scanf
#define pf printf
#define pb push_back
#define mp make_pair
#define PI ( acos(-1.0) )
#define mod 1000000007
#define maxn 100005
#define IN freopen("nocross.in","r",stdin)
#define OUT freopen("nocross.out","w",stdout)
#define FOR(i,a,b) for(i=a ; i<=b ; i++)
#define DBG pf("Hi\n")
#define INF 2000000000
#define i64 long long int
#define eps (1e-8)
#define xx first
#define yy second
#define ln 17
#define off 2

// using namespace __gnu_pbds;
using namespace std ;

// typedef tree< i64, null_type, less<i64>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;

char s[105][105] ;
int A , B ;

int main()
{
    int i , j , k , l , m , n ;

    scanf("%d %d",&A,&B) ;

    for(i=0 ; i<100 ; i++)
    {
        s[0][i] = 'B' ; s[49][i] = 'B' ;
        s[50][i] = 'W' ; s[99][i] = 'W' ;
    }

    for(i=0 ; i<50 ; i++)
    {
        s[i][0] = 'B' ; s[i][99] = 'B' ;
        s[50+i][0]='W' ; s[50+i][99] = 'W' ;
    }

    A-- ; B-- ;

    for(i=1 ; i<49 ; i++)
    {
        if(i%2==0)
        {
            for(j=1 ; j<99 ; j++) s[i][j] = 'B' ;
            continue ;
        }
        for(j=1 ; j<99 ; j++)
        {
            if( A==0 ) s[i][j] = 'B' ;
            else
            {
                if( s[i-1][j]=='B' && s[i][j-1]=='B' )
                {
                    s[i][j] = 'W' ;
                    A-- ;
                }
                else s[i][j] = 'B' ;
            }
        }
    }

    for(i=51 ; i<99 ; i++)
    {
        if(i%2==0){
            for(j=1 ; j<99 ; j++) s[i][j] = 'W' ;
            continue ;
        }
        for(j=1 ; j<99 ; j++)
        {
            if( B==0 ) s[i][j] = 'W' ;
            else
            {
                if( s[i-1][j]=='W' && s[i][j-1]=='W' )
                {
                    s[i][j] = 'B' ;
                    B-- ;
                }
                else s[i][j] = 'W' ;
            }
        }
    }

    printf("100 100\n") ;

    for(i=0 ; i<100 ; i++)
    {
        for(j=0 ; j<100 ; j++)
        {
            if( s[i][j]=='B' ) s[i][j] = '#' ;
            else s[i][j] = '.' ;
        }
    }

    for(i=0 ; i<100 ; i++)
    {
        s[i][100] = '\0' ;
        printf("%s\n",s[i]) ;
    }

    return 0 ;
}