Problem: 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.

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 `#`.

Examples

Input

2 3


Output

3 3
##.
..#
#.#


Input

7 8


Output

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


Input

1 1


Output

4 2
..
.


Input

3 14


Output

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

Solution (in CPP):

Solution:
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
typedef long long  ll;
#define rep(i,s,n)for(ll i=s;i<n;i++)
#define repe(i,s,n)for(ll i=s;i<=n;i++)

int main()
{
	ll a, b; cin >> a >> b;
	ll s = 100;
	string m[100];
	rep(i, 0, s)m[i] = string(s, ((i < 50) ? '.' : '#'));
	b--, a--;

	rep(i, 0, s) {
		if (!b)break;
		rep(j, 0, s) {
			if (!b)break;
			if (!(i & 1) && j & 1) {
				m[i][j] = '#';
				b--;
			}
		}
	}

	rep(i, s / 2 + 1, s) {
		if (!a)break;
		rep(j, 0, s) {
			if (!a)break;
			if (!(i & 1) && j & 1) {
				m[i][j] = '.';
				a--;
			}
		}
	}

	cout << 100 << " " << 100 << endl;
	rep(i, 0, s) {
		rep(j, 0, s)cout << m[i][j];
		cout << endl;
	}
	return 0;
}