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;
long long n, k, l, i, j, l1, nr, cate[100][100];
char m[40][2000];
int main() {
  cin >> n >> k;
  for (i = 1; i <= 4; i++)
    for (j = 1; j <= n; j++) m[i][j] = '.';
  if (k == n - 2) {
    for (j = 2; j < n; j++) m[2][j] = '#';
  } else if (k % 2 == 0) {
    for (j = 2; j <= 1 + (k / 2); j++) m[2][j] = m[3][j] = '#';
  } else if (k > 3) {
    k++;
    for (j = 2; j <= 1 + (k / 2); j++) m[2][j] = m[3][j] = '#';
    for (j = 3; j < n - 1; j++)
      if (m[2][j] == '#' && m[2][j - 1] == '#' && m[2][j + 1] == '#') {
        m[2][j] = '.';
        j = n + 10;
      }
    k--;
  } else if (k == 3) {
    m[2][2] = '#';
    m[2][n - 1] = '#';
    m[3][(n + 1) / 2] = '#';
  } else if (k == 1) {
    m[2][(n + 1) / 2] = '#';
  }
  nr = 0;
  for (i = 2; i <= 3; i++)
    for (j = 1; j <= n; j++)
      if (m[i][j] == '#') nr++;
  if (nr == k) {
    cout << "YES\n";
    for (i = 1; i <= 4; i++) {
      for (j = 1; j <= n; j++) cout << m[i][j];
      cout << "\n";
    }
    cate[0][1] = 1;
    for (i = 1; i <= 4; i++)
      for (j = 1; j <= n; j++)
        if (m[i][j] == '.') cate[i][j] = cate[i][j - 1] + cate[i - 1][j];
    cate[5][1] = 1;
    for (i = 4; i >= 1; i--)
      for (j = 1; j <= n; j++)
        if (m[i][j] == '.') cate[i][j] = cate[i][j - 1] + cate[i + 1][j];
  } else
    cout << "NO\n";
}
