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 int power(long long int x, long long int y);
long long int gcd(long long int a, long long int b);
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, k;
  cin >> n >> k;
  if (k % 2 == 0) {
    cout << "YES\n";
    for (long long int i = 0; i < n; i++) cout << '.';
    cout << "\n";
    long long int temp = k / 2;
    cout << '.';
    for (long long int i = 0; i < temp; i++) cout << '#';
    for (long long int i = 0; i < (n - 2 - temp); i++) cout << '.';
    cout << '.' << '\n';
    cout << '.';
    for (long long int i = 0; i < temp; i++) cout << '#';
    for (long long int i = 0; i < (n - 2 - temp); i++) cout << '.';
    cout << '.' << '\n';
    for (long long int i = 0; i < n; i++) cout << '.';
    cout << "\n";
  } else {
    long long int temp = n - 2;
    cout << "YES\n";
    for (long long int i = 0; i < n; i++) cout << '.';
    cout << "\n";
    cout << '.';
    if (k <= temp) {
      for (long long int i = 0; i < (temp - k) / 2; i++) cout << '.';
      for (long long int i = 0; i < k; i++) cout << '#';
      for (long long int i = 0; i < (temp - k) / 2; i++) cout << '.';
    } else
      for (long long int i = 0; i < temp; i++) cout << '#';
    cout << '.' << '\n';
    if (k > temp) {
      k -= temp;
      k -= 2;
      cout << '.';
      cout << '#';
      for (long long int i = 0; i < k; i++) cout << '#';
      for (long long int i = 0; i < (temp - k - 2); i++) cout << '.';
      cout << '#';
      cout << '.' << '\n';
    } else {
      for (long long int i = 0; i < n; i++) cout << '.';
      cout << "\n";
    }
    for (long long int i = 0; i < n; i++) cout << '.';
    cout << "\n";
  }
}
long long int power(long long int x, long long int y) {
  long long int res = 1;
  x %= 1000000007;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y >> 1;
    x = (x * x) % 1000000007;
  }
  return res % 1000000007;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
