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;
int n, k;
int m[2000][2000];
void fillFirstLine() {
  int x, y;
  y = 1;
  int l = n / 2, r = n / 2;
  while ((l > 0) && (k > 0)) {
    if (l == r) {
      m[y][l] = 1;
      k--;
    } else {
      m[y][l] = 1;
      m[y][r] = 1;
      k -= 2;
    }
    l--;
    r++;
  }
}
void fillSecondLine() {
  int x, y;
  y = 2;
  int l = 1, r = n - 2;
  while (k > 0) {
    if (l == r) {
      m[y][l] = 1;
      k--;
    } else {
      m[y][l] = 1;
      m[y][r] = 1;
      k -= 2;
    }
    l++;
    r--;
  }
}
void printAns() {
  cout << "YES" << endl;
  for (int i = 0; i < 4; i++) {
    for (int j = 0; j < n; j++) {
      if (m[i][j] == 0) cout << '.';
      if (m[i][j] == 1) cout << '#';
    }
    cout << endl;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  if (k == 0) {
    printAns();
    return 0;
  }
  if (k % 2 == 0) {
    for (int i = 0 + 1; i < n - 1; i++) {
      for (int j = 0 + 1; j < 4 - 1; j++) {
        k--;
        m[j][i] = 1;
        if (k <= 0) {
          printAns();
          return 0;
        }
      }
    }
  }
  fillFirstLine();
  if (k > 0) {
    fillSecondLine();
  }
  printAns();
  return 0;
}
