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 main() {
  int n, k;
  cin >> n >> k;
  if (k % 2 == 0 && k / 2 <= n - 2) {
    cout << "YES" << endl;
    vector<vector<char> > v(4, vector<char>(n, '.'));
    int i = 1;
    while (k) {
      v[1][i] = '#';
      v[2][i] = '#';
      i++;
      k = k - 2;
    }
    for (long long i = 0; i < 4; i++) {
      for (long long j = 0; j < n; j++) {
        cout << v[i][j];
      }
      cout << endl;
    }
  } else if (k % (n - 2) == 0 && k / (n - 2) <= 2) {
    cout << "YES" << endl;
    vector<vector<char> > v(4, vector<char>(n, '.'));
    int j = 1;
    while (k) {
      for (int i = 1; i < n - 1; i++) {
        v[j][i] = '#';
      }
      j++;
      k = k - (n - 2);
    }
    for (long long i = 0; i < 4; i++) {
      for (long long j = 0; j < n; j++) {
        cout << v[i][j];
      }
      cout << endl;
    }
  } else if (k % 2 != 0 && k < 2 * (n - 2)) {
    cout << "YES" << endl;
    vector<vector<char> > v(4, vector<char>(n, '.'));
    v[1][n / 2] = '#';
    k--;
    long long i = 1, j = n - 2;
    while (k && i < j) {
      v[1][i] = '#';
      v[1][j] = '#';
      k -= 2;
      i++;
      j--;
    }
    i = 1, j = n - 2;
    while (k > 0 && i <= j) {
      v[2][i] = '#';
      v[2][j] = '#';
      k -= 2;
      i++;
      j--;
    }
    for (long long i = 0; i < 4; i++) {
      for (long long j = 0; j < n; j++) {
        cout << v[i][j];
      }
      cout << endl;
    }
  } else if (k < n - 2) {
    cout << "YES" << endl;
    vector<vector<char> > v(4, vector<char>(n, '.'));
    for (long long i = 1, j = 0; i < n - 1 && j < k; i++, j++) {
      v[1][i] = '#';
    }
    for (long long i = 0; i < 4; i++) {
      for (long long j = 0; j < n; j++) {
        cout << v[i][j];
      }
      cout << endl;
    }
  } else
    cout << "NO" << endl;
  return 0;
}
