description: The only difference between this problem and D1 is that you don't have to provide the way to construct the answer in D1, but you have to do it in this problem.
  
  There's a table of n × m cells (n rows and m columns). The value of n ⋅ m is even.
  
  A domino is a figure that consists of two cells having a common side. It may be horizontal (one of the cells is to the right of the other) or vertical (one of the cells is above the other).
  
  You need to place nm/2 dominoes on the table so that exactly k of them are horizontal and all the other dominoes are vertical. The dominoes cannot overlap and must fill the whole table.
  
  Input
  
  The first line contains one integer t (1 ≤ t ≤ 10) — the number of test cases. Then t test cases follow.
  
  Each test case consists of a single line. The line contains three integers n, m, k (1 ≤ n,m ≤ 100, 0 ≤ k ≤ nm/2, n ⋅ m is even) — the count of rows, columns and horizontal dominoes, respectively.
  
  Output
  
  For each test case:
  
    * print "NO" if it's not possible to place the dominoes on the table in the described way; 
    * otherwise, print "YES" on a separate line, then print n lines so that each of them contains m lowercase letters of the Latin alphabet — the layout of the dominoes on the table. Each cell of the table must be marked by the letter so that for every two cells having a common side, they are marked by the same letters if and only if they are occupied by the same domino. I.e. both cells of the same domino must be marked with the same letter, but two dominoes that share a side must be marked with different letters. If there are multiple solutions, print any of them. 
  
  Example
  
  Input
  
  
  8
  4 4 2
  2 3 0
  3 2 3
  1 2 0
  2 4 2
  5 2 2
  2 17 16
  2 1 1
  
  
  Output
  
  
  YES
  accx
  aegx
  bega
  bdda
  YES
  aha
  aha
  YES
  zz
  aa
  zz
  NO
  YES
  aaza
  bbza
  NO
  YES
  bbaabbaabbaabbaay
  ddccddccddccddccy
  NO
solution: oa = ord('a')
  def slv(n, m, k):
      ans = [[-1]*m for i in range(n)]
      # assume n even
      if k & 1:
          return None
      if k > (m//2)*n:
          return None
      cnt = 0
      for i in range(0,n,2):
          for j in range(0,m-1,2):
              hh = ((i//2)%2)*2+((j//2)%2)
              c1 = 2*hh
              c2 = 2*hh+1
              if cnt == k:
                  #vert
                  ans[i][j] = c1
                  ans[i+1][j] = c1
                  ans[i][j+1] = c2
                  ans[i+1][j+1] = c2
              else:
                  ans[i][j] = c1
                  ans[i][j+1] = c1
                  ans[i+1][j] = c2
                  ans[i+1][j+1] = c2
                  cnt += 2
      if m & 1:
          for j in range(0, n, 2):
              c = 8 if (j&2) else 9
              ans[j][m-1] = c
              ans[j+1][m-1] = c
      
      return ans
  
  
  
  
  
  for _ in range(int(input())):
      n, m, k = [int(i) for i in input().split()]
      ans = []
      if n & 1:
          if k >= (m//2):
              ans = slv(n-1,m,k-(m//2))
              if ans is None:
                  print('NO')
                  continue
              lst = [-1]*m
              for i in range(0,m,2):
                  c = 10 if (i & 2) else 11
                  lst[i] = c
                  lst[i+1] = c
              ans.append(lst)
          else:
              print('NO')
              continue
      else:
          ans = slv(n,m,k)
          if ans is None:
              print('NO')
              continue
      
      print('YES')
      for rw in ans:
          print(''.join(chr(p+oa) for p in rw))
  