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() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, k;
  cin >> n >> k;
  long long q = k % 2;
  long long r = k % (n - 2);
  if (q == 0 && (k / 2) <= (n - 2)) {
    cout << "YES" << endl;
    for (int i = 0; i < 4; i++) {
      long long rem = k / 2;
      for (int j = 0; j < n; j++) {
        if (i == 0 || i == 3 || j == 0 || j == n - 1) {
          cout << ".";
        } else {
          if (rem) {
            cout << "#";
            rem--;
          } else {
            cout << ".";
          }
        }
      }
      cout << endl;
    }
    return 0;
  } else if (r == 0 && k / (n - 2) <= 2) {
    cout << "YES" << endl;
    long long rem = k;
    for (int i = 0; i < 4; i++) {
      for (int j = 0; j < n; j++) {
        if (i == 0 || i == 3 || j == 0 || j == n - 1) {
          cout << ".";
        } else {
          if (rem) {
            cout << "#";
            rem--;
          } else {
            cout << ".";
          }
        }
      }
      cout << endl;
    }
    return 0;
  } else if (n >= 5 && k >= (n - 2) + 2) {
    cout << "YES" << endl;
    long long rem = k;
    for (int i = 0; i < 2; i++) {
      for (int j = 0; j < n; j++) {
        if (i == 0 || i == 3 || j == 0 || j == n - 1) {
          cout << ".";
        } else if (i == 1) {
          cout << "#";
          k--;
        }
      }
      cout << endl;
    }
    k--;
    for (int i = 0; i < 2; i++) {
      for (int j = 0; j < n; j++) {
        if (i == 1 || j == 0 || j == n - 1) {
          cout << ".";
        } else if (j == n - 2) {
          cout << "#";
        } else {
          if (k) {
            cout << "#";
            k--;
          } else
            cout << ".";
        }
      }
      cout << endl;
    }
    return 0;
  } else if (k & 1) {
    cout << "YES" << endl;
    char dp[4][1000];
    for (int i = 0; i < 4; i++) {
      for (int j = 0; j < n; j++) {
        dp[i][j] = '.';
      }
    }
    int p = 0;
    for (int i = n / 2; i >= 0 && k > 0; i--) {
      dp[(p % 2) + 1][i] = '#';
      if (dp[(p % 2) + 1][n - i - 1] == '.') {
        dp[p % 2 + 1][n - i - 1] = '#';
        k--;
      }
      k--;
      p++;
    }
    for (int i = 0; i < 4; i++) {
      for (int j = 0; j < n; j++) {
        cout << dp[i][j];
      }
      cout << endl;
    }
    return 0;
  }
  cout << "NO" << endl;
  return 0;
  return 0;
}
