The city of Fishtopia can be imagined as a grid of 4 rows and an odd number of columns. It has two main villages; the first is located at the top-left cell (1,1), people who stay there love fishing at the Tuna pond at the bottom-right cell (4, n). The second village is located at (4, 1) and its people love the Salmon pond at (1, n).

The mayor of Fishtopia wants to place k hotels in the city, each one occupying one cell. To allow people to enter the city from anywhere, hotels should not be placed on the border cells.

A person can move from one cell to another if those cells are not occupied by hotels and share a side.

Can you help the mayor place the hotels in a way such that there are equal number of shortest paths from each village to its preferred pond?

Input

The first line of input contain two integers, n and k (3 ≤ n ≤ 99, 0 ≤ k ≤ 2×(n-2)), n is odd, the width of the city, and the number of hotels to be placed, respectively.

Output

Print "YES", if it is possible to place all the hotels in a way that satisfies the problem statement, otherwise print "NO".

If it is possible, print an extra 4 lines that describe the city, each line should have n characters, each of which is "#" if that cell has a hotel on it, or "." if not.

Examples

Input

7 2


Output

YES
.......
.#.....
.#.....
.......


Input

5 3


Output

YES
.....
.###.
.....
.....
#include <bits/stdc++.h>
using namespace std;
bool Less(const pair<pair<int, int>, int>& s1,
          const pair<pair<int, int>, int>& s2) {
  if (s1.first.first == s2.first.first)
    return s1.first.second > s2.first.second;
  return s1.first.first < s2.first.first;
}
int main() {
  int n, k;
  cin >> n >> k;
  char da[4][200];
  if (n >= 5 && k >= 5)
    cout << "YES" << endl;
  else if (k == 1 || k == 3)
    cout << "YES" << endl;
  else if (k % 2 == 0 || k % (n - 2) == 0)
    cout << "YES" << endl;
  else {
    cout << "NO" << endl;
    return 0;
  }
  if (k % 2 == 0)
    for (int j = 0; j < n; j++) {
      for (int i = 0; i < 4; i++)
        if (k != 0 && j != 0 && j != n - 1 && i != 0 && i != 3)
          da[i][j] = '#', k--;
        else
          da[i][j] = '.';
    }
  else if (k % (n - 2) == 0)
    for (int i = 0; i < 4; i++) {
      for (int j = 0; j < n; j++)
        if (k != 0 && j != 0 && j != n - 1 && i != 0 && i != 3)
          da[i][j] = '#', k--;
        else
          da[i][j] = '.';
    }
  else if (n >= 5 && k >= 5) {
    k = k - 2;
    for (int j = 0; j < n; j++) {
      for (int i = 0; i < 4; i++)
        if (k != 0 && j != 0 && j != n - 1 && i != 0 && i != 3)
          da[i][j] = '#', k--;
        else
          da[i][j] = '.';
    }
    for (int j = 1; j < n; j++)
      if (da[1][j] == '.') {
        da[1][j] = da[2][j] = '#';
        break;
      }
  } else if (k == 1 || k == 3) {
    for (int i = 0; i < 4; i++)
      for (int j = 0; j < n; j++) da[i][j] = '.';
    int j = n / 2;
    if (k == 1)
      da[1][j] = '#';
    else
      da[1][j - 1] = '#', da[1][j] = '#', da[1][j + 1] = '#';
  }
  for (int i = 0; i < 4; i++) {
    for (int j = 0; j < n; j++) cout << da[i][j];
    cout << endl;
  }
  return 0;
}
