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 isPowerOfTwo(long long n) {
  if (n == 0) return false;
  return (ceil(log2(n)) == floor(log2(n)));
}
long long power(long long x, unsigned long long y) {
  if (y == 0)
    return 1;
  else if (y % 2 == 0)
    return power(x, y / 2) * power(x, y / 2);
  else
    return x * power(x, y / 2) * power(x, y / 2);
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long phi(long long n) {
  long long result = n;
  for (long long p = 2; p * p <= n; ++p) {
    if (n % p == 0) {
      while (n % p == 0) n /= p;
      result -= result / p;
    }
  }
  if (n > 1) result -= result / n;
  return result;
}
signed main() {
  long long n, k;
  cin >> n >> k;
  char a[4][n];
  for (long long i = 0; i < 4; i++) {
    for (long long j = 0; j < n; j++) a[i][j] = '.';
  }
  if (k % 2 == 0) {
    long long j = 1;
    while (k > 0) {
      a[1][j] = '#';
      a[2][j] = '#';
      j++;
      k -= 2;
    }
  } else {
    a[1][n / 2] = '#';
    k--;
    long long j = 1;
    while (k > 0 && j < (n - 1) / 2) {
      a[1][j] = '#';
      a[1][n - 1 - j] = '#';
      j++;
      k -= 2;
    }
    j = 1;
    while (k > 0) {
      a[2][j] = '#';
      a[2][n - 1 - j] = '#';
      j++;
      k -= 2;
    }
  }
  cout << "YES" << endl;
  for (long long i = 0; i < 4; i++) {
    for (long long j = 0; j < n; j++) cout << a[i][j];
    cout << endl;
  }
  return 0;
}
