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 col, ara[200][200], blk;
void printLn() {
  for (int i = 0; i < col; i++) cout << ".";
  cout << endl;
}
void printEv(int tmp) {
  cout << ".";
  for (int i = 0; i < tmp; i++) cout << "#";
  for (int i = tmp + 1; i < col; i++) cout << ".";
  cout << endl;
}
void printOdd(int tmp) {
  for (int i = 0; i < (col / 2 - tmp / 2); i++) cout << ".";
  for (int i = 0; i < (tmp / 2); i++) cout << "#";
  cout << "#";
  for (int i = 0; i < (tmp / 2); i++) cout << "#";
  for (int i = 0; i < (col / 2 - tmp / 2); i++) cout << ".";
  cout << endl;
}
void printOdd1(int tmp) {
  cout << ".";
  for (int i = 0; i < tmp / 2; i++) cout << "#";
  for (int i = tmp / 2 + 1; i < col / 2; i++) cout << ".";
  cout << ".";
  for (int i = tmp / 2 + 1; i < col / 2; i++) cout << ".";
  for (int i = 0; i < tmp / 2; i++) cout << "#";
  cout << "." << endl;
}
int main() {
  cin >> col >> blk;
  if (blk % 2 == 0) {
    cout << "YES" << endl;
    printLn();
    printEv(blk / 2);
    printEv(blk / 2);
    printLn();
  } else {
    cout << "YES" << endl;
    printLn();
    printOdd(min(blk, col - 2));
    blk = max(0, blk - (col - 2));
    printOdd1(blk);
    printLn();
  }
  return 0;
}
