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 pwr(int base, int p) {
  int ans = 1;
  while (p) {
    if (p & 1) ans = (ans * base) % 1000000007;
    base = (base * base) % 1000000007;
    p /= 2;
  }
  return ans;
}
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int powr(int base, int p) {
  int ans = 1;
  while (p) {
    if (p & 1) ans = (ans * base);
    base = (base * base);
    p /= 2;
  }
  return ans;
}
int pre[200005];
int search(int l, int h, int val2) {
  int ans1 = -1;
  while (l <= h) {
    int m = (l + h) / 2;
    if (pre[m] > val2) {
      ans1 = m;
      h = m - 1;
    } else
      l = m + 1;
  }
  return ans1;
}
struct node {
  int x, y, i;
};
bool comp(node a, node b) { return a.y > b.y; }
vector<int> g[200005];
int main() {
  int n, k;
  scanf("%d", &n);
  scanf("%d", &k);
  char s[4][n];
  cout << "YES" << endl;
  for (int i = 0; i < 4; i++) {
    for (int j = 0; j < n; j++) {
      s[i][j] = 'a';
    }
  }
  for (int i = 1; i <= 2; i++) {
    for (int j = 1; j < n / 2; j++) {
      if (k >= 2) {
        s[i][j] = '#';
        s[i][n - j - 1] = '#';
        k -= 2;
      } else
        break;
    }
  }
  if (k == 1) {
    s[1][n / 2] = '#';
  }
  if (k == 2) {
    s[1][n / 2] = '#';
    s[2][n / 2] = '#';
  }
  for (int i = 0; i < 4; i++) {
    for (int j = 0; j < n; j++) {
      if (s[i][j] == 'a') s[i][j] = '.';
      cout << s[i][j];
    }
    cout << endl;
  }
}
