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;
const int MX = 200000;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int lcm(int x, int y) {
  x /= gcd(x, y);
  return x * y;
}
bool prime[MX + 5], isprime[MX + 5];
vector<int> v;
int cs = 0;
int fx[] = {1, -1, 0, 0};
int fy[] = {0, 0, 1, -1};
set<int> st;
set<int>::iterator it;
char grid[305][305];
bool InRng(int i, int j, int r, int c) {
  return ((i >= 0 && i < r) && (j >= 0 && j < c));
}
int ar[1234][1234];
long long n, m, k, s, x, y;
long long a[25], b[25];
void solve() {
  int n, k;
  cin >> n >> k;
  if (k % 2 == 0) {
    int c = 0;
    for (int i = 2;; i++) {
      if (c == k) break;
      ar[2][i] = 1;
      ar[3][i] = 1;
      c += 2;
    }
  } else {
    int c = 0;
    int l = n / 2 + n % 2;
    int r = l;
    while (c < k) {
      ar[2][l] = 1;
      c++;
      l--;
      if (r != n / 2 + n % 2) {
        ar[2][r] = 1;
        c++;
      }
      r++;
      if (l == 1) break;
    }
    l = n / 2 + n % 2 - 1;
    r = n / 2 + n % 2 + 1;
    while (c < k) {
      ar[3][l] = 1;
      l--;
      c++;
      ar[3][r] = 1;
      r++;
      c++;
    }
  }
  puts("YES");
  for (int i = 1; i < 5; i++) {
    for (int j = 1; j <= n; j++) {
      if (ar[i][j])
        cout << "#";
      else
        cout << ".";
    }
    putchar(10);
  }
}
void sieve() {
  v.push_back(2);
  prime[1] = 1;
  int limit = (int)sqrt(MX);
  for (int i = 4; i <= MX; i += 2) prime[i] = 1;
  for (int i = 3; i <= MX; i += 2) {
    if (!prime[i]) {
      v.push_back(i);
      if (i <= limit) {
        for (int j = i * i; j <= MX; j += 2 * i) prime[j] = 1;
      }
    }
  }
}
int main() {
  solve();
  return 0;
}
