description: A binary string is a string that consists of characters 0 and 1.
  
  Let \operatorname{MEX} of a binary string be the smallest digit among 0, 1, or 2 that does not occur in the string. For example, \operatorname{MEX} of 001011 is 2, because 0 and 1 occur in the string at least once, \operatorname{MEX} of 1111 is 0, because 0 and 2 do not occur in the string and 0 < 2.
  
  A binary string s is given. You should cut it into any number of substrings such that each character is in exactly one substring. It is possible to cut the string into a single substring — the whole string.
  
  A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
  
  What is the minimal sum of \operatorname{MEX} of all substrings pieces can be?
  
  Input
  
  The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.
  
  Each test case contains a single binary string s (1 ≤ |s| ≤ 10^5).
  
  It's guaranteed that the sum of lengths of s over all test cases does not exceed 10^5.
  
  Output
  
  For each test case print a single integer — the minimal sum of \operatorname{MEX} of all substrings that it is possible to get by cutting s optimally.
  
  Example
  
  Input
  
  
  6
  01
  1111
  01100
  101
  0000
  01010
  
  
  Output
  
  
  1
  0
  2
  1
  1
  2
  
  Note
  
  In the first test case the minimal sum is \operatorname{MEX}(0) + \operatorname{MEX}(1) = 1 + 0 = 1.
  
  In the second test case the minimal sum is \operatorname{MEX}(1111) = 0.
  
  In the third test case the minimal sum is \operatorname{MEX}(01100) = 2.
solution: import sys
  testing = len(sys.argv) == 4 and sys.argv[3] == "myTest"
  interactive = False
  if testing:
      cmd = sys.stdout
      from time import time
      start_time = int(round(time() * 1000)) 
      readAll = open(sys.argv[1], 'r').read
      sys.stdout = open(sys.argv[2], 'w')
  else:
      readAll = sys.stdin.read
  
  # ############ ---- I/O Functions ---- ############
  
  class InputData:
      def __init__(self):
          self.lines = readAll().split('\n')
          self.n = len(self.lines)
          self.ii = -1
      def input(self):
          self.ii += 1
          assert self.ii < self.n
          return self.lines[self.ii]
  
  flush = sys.stdout.flush
  if interactive and not testing:
      input = sys.stdin.readline
  else:
      inputData = InputData()
      input = inputData.input
  
  def intin():
      return(int(input()))
  def intlin():
      return(list(map(int,input().split())))
  def chrin():
      return(list(input()))
  def strin():
      return input()
  def lout(l, sep="\n", toStr=True):
      print(sep.join(map(str, l) if toStr else l))
  def dout(*args, **kargs):
      if not testing: return
      if args: print(args[0] if len(args)==1 else args)
      if kargs: print([(k,v) for k,v in kargs.items()])
  def ask(q):
      sys.stdout.write(str(q)+'\n')
      flush()
      return intin()
      
  # ############ ---- I/O Functions ---- ############
  
  # from math import ceil
  # from collections import defaultdict as ddict, Counter
  # from heapq import *
  # from Queue import Queue
  
  def main():
      s = strin()
      ss = set(s)
      if len(ss) == 1:
          if '0' in ss:
              return 1
          return 0
      flag = False
      i = 0
      n = len(s)
      while i < n:
          if s[i] == '0':
              if flag:
                  return 2
              while i < n and s[i] == '0':
                  i += 1
              flag = True
          else:
              i += 1
      return 1
          
  
  anss = []
  for _ in xrange(intin()):
      anss.append(main())
      # anss.append("YES" if main() else "NO")
  lout(anss)
  
  if testing:
      sys.stdout = cmd
      print(int(round(time() * 1000))  - start_time)