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;
struct S {
  int a, b;
  S() {}
  S(int _a, int _b) {
    a = _a;
    b = _b;
  }
  const bool operator<(const S &o) const { return a < o.a; }
};
priority_queue<int, vector<int>, greater<int>> mpq;
inline void showAll(vector<int> &v, int NL_or_space) {
  for (int &here : v) {
    printf("%d", here);
    printf("%c", (NL_or_space) ? '\n' : ' ');
  }
}
const int SIZE = 109;
char grid[SIZE][SIZE];
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= 4; i++)
    for (int j = 1; j <= n; j++) grid[i][j] = '.';
  int ans = -1;
  if (k == 0)
    ans = 1;
  else if (k % 2 == 0) {
    ans = 1;
    for (int i = 0; i < k / 2; i++) {
      grid[2][2 + i] = '#';
      grid[3][2 + i] = '#';
    }
  } else {
    ans = 1;
    grid[2][(1 + n) / 2] = '#';
    k--;
    for (int i = 1; i <= k / 2; i++) {
      grid[2][(1 + n) / 2 + i] = '#';
      grid[2][(1 + n) / 2 - i] = '#';
      k -= 2;
    }
    if (k & 1) assert(0);
    for (int i = 0; i < k / 2; i++) {
      grid[3][i + 2] = '#';
      grid[3][n - 1 - i] = '#';
    }
  }
  if (ans) {
    printf("YES\n");
    for (int i = 1; i <= 4; i++) {
      for (int j = 1; j <= n; j++) {
        printf("%c", grid[i][j]);
      }
      printf("\n");
    }
  } else
    printf("NO\n");
  return 0;
}
