def r(a):
 i=a.find('0')
 if i<0:
     print a
 [m in[(i-j)%9*(i/9^j/9)*(i/27^j/27|i%9/3^j%9/3)or a[j]for
 j in range(81)]or r(a[:i]+m+a[i+1:])for m in`14**7*9`]
 r(raw_input())

#solve sudoku by backtracking
def solve_sudoku(board):
 
 #i is the first non-filled place, and the one we're going to try different values for
 i=board.find('0')
 
 #if i == -1, the puzzle is solved
 if i<0:
     print board
 
 #iterate over all squares j; if the big test returns true, then j is in the same column, block or row as i and needs to be considered
 #if the test is true, the value of j square is appended, meaning the final result is a list of all the values the i square surely cannot have
 
 # python has a nice feature called lazy evaluation. 
 # if the first clause is true, the result of the other one is irrelevant for the output of "or". 
 # therefore we use "a or b" to execute b only if a is false.
 exclusion_list = [(i-j)%9*(i/9^j/9)*(i/27^j/27|i%9/3^j%9/3)or a[j] for j in range(81)]
 
#14**7*9 is 948721536, which contains all the digits from 1-9.
#therefore "for m in `14**7*9` and the below are equivalent.
#iterate through all the possible values for this square i.
 for m in "123456789":
 #lazy evaluation, again!
 #recursively call solve_sudoku itself with board[i] = m if this value of m is not on our exclusion list.
  [m in exclusion_list or solve_sudoku(board[:i]+m+board[i+1:])]
 
#solve sudoku puzzle
solve_sudoku(raw_input())