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 dp[200008], arr[200008];
int main() {
  long long n, m, k, a, b;
  long long sum = 0;
  string s;
  cin >> n >> k;
  a = n - 2;
  if (k % a == 0 && k <= (a * 2)) {
    cout << "YES" << endl;
    for (int i = 0; i < 4; i++) {
      for (int j = 0; j < n; j++) {
        if (i == 0 || j == 0 || i == 3 || j == n - 1) {
          cout << ".";
        } else {
          if (k) {
            cout << "#";
            k--;
          } else
            cout << ".";
        }
      }
      cout << endl;
    }
    return 0;
  }
  if (k % 2 == 0 && k <= (a * 2)) {
    int g = k / 2;
    cout << "YES" << endl;
    for (int i = 0; i < 4; i++) {
      int p = g;
      for (int j = 0; j < n; j++) {
        if (i == 0 || j == 0 || i == 3 || j == n - 1) {
          cout << ".";
        } else {
          if (p) {
            cout << "#";
            p--;
          } else
            cout << ".";
        }
      }
      cout << endl;
    }
    return 0;
  }
  if ((k % a) % 2 == 1 && k <= (a * 2)) {
    cout << "YES" << endl;
    char kp[101][101];
    memset((kp), ('.'), sizeof(kp));
    if (k / a == 1) {
      for (int i = 1; i < n - 1; i++) {
        kp[1][i] = '#';
      }
      k = k % a;
      int f = (a - k) / 2;
      for (int i = (1 + f); i < (k + 1 + f); i++) {
        kp[2][i] = '#';
      }
    } else {
      k = k % a;
      int f = (a - k) / 2;
      for (int i = (1 + f); i < (k + 1 + f); i++) {
        kp[1][i] = '#';
      }
    }
    for (int i = 0; i < 4; i++) {
      for (int j = 0; j < n; j++) cout << kp[i][j];
      cout << endl;
    }
    return 0;
  }
  if ((k % a) % 2 == 0 && k <= (a * 2)) {
    cout << "YES" << endl;
    char kp[101][101];
    memset((kp), ('.'), sizeof(kp));
    if (k / a == 1) {
      for (int i = 1; i < n - 1; i++) {
        kp[1][i] = '#';
      }
      k = k % a;
      int f = (a - k) / 2;
      int ff = (n) / 2;
      for (int i = (1 + f); i <= (k + 1 + f); i++) {
        if (i != ff) kp[2][i] = '#';
      }
    } else {
      k = k % a;
      int f = (a - k) / 2;
      int ff = (n) / 2;
      for (int i = (1 + f); i <= (k + 1 + f); i++) {
        if (ff != i) kp[1][i] = '#';
      }
    }
    for (int i = 0; i < 4; i++) {
      for (int j = 0; j < n; j++) cout << kp[i][j];
      cout << endl;
    }
    return 0;
  }
  cout << "NO" << endl;
  return 0;
}
