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>

#define each(i, c) for (auto& i : c)
#define unless(cond) if (!(cond))

using namespace std;

typedef long long int lli;
typedef unsigned long long ull;
typedef complex<double> point;

template<typename P, typename Q>
ostream& operator << (ostream& os, pair<P, Q> p)
{
  os << "(" << p.first << "," << p.second << ")";
  return os;
}

template<typename T>
ostream& operator << (ostream& os, vector<T> v)
{
  os << "(";
  each (i, v) os << i << ",";
  os << ")";
  return os;
}

int main(int argc, char *argv[])
{
  ios_base::sync_with_stdio(0);
  cin.tie(0);

  int a, b;
  while (cin >> a >> b) {
    swap(a, b);
    
    const int N = 100;
    int g[N][N];
    fill(&g[0][0], &g[N - 1][N - 1] + 1, '@');

    vector<pair<int, int>> v;
    vector<pair<int, int>> u;

    set<pair<int, int>> taboo;
    taboo.insert(make_pair(0, 0));
    taboo.insert(make_pair(N-1, N-1));
    taboo.insert(make_pair(0, N-1));
    taboo.insert(make_pair(N-1, 0));
    
    int di[] = {0, +1, 0, -1};
    int dj[] = {+1, 0, -1, 0};
    {
      int dir = 0;
      int len = N;
      int i = 0;
      int j = -1;
      while (len) {
        for (int k = 0; k < len; ++k) {
          i += di[dir];
          j += dj[dir];
          g[i][j] = '#';
          unless (taboo.count(make_pair(i, j))) v.push_back(make_pair(i, j));
        }
        dir = (dir + 1) % 4;
        --len;
      }
    }

    {
      int dir = 3;
      int len = N - 1;
      int i = N;
      int j = 0;
      while (len) {
        for (int k = 0; k < len; ++k) {
          i += di[dir];
          j += dj[dir];
          g[i][j] = '.';
          unless (taboo.count(make_pair(i, j))) u.push_back(make_pair(i, j));
        }
        dir = (dir + 1) % 4;
        --len;
      }
    }

    reverse(u.begin(), u.end());

    for (int i = 0; i < a - 1; ++i) {
      int j = 2 * i + 1;
      g[v[j].first][v[j].second] = '.';
    }
    for (int i = 0; i < b - 1; ++i) {
      int j = 2 * i + 1;
      g[u[j].first][u[j].second] = '#';
    }

    cout << N << ' ' << N << endl;
    for (int i = 0; i < N; ++i) {
      for (int j = 0; j < N; ++j) {
        cout << (char)g[i][j];
      }
      cout << endl;
    }
  }
  
  return 0;
}