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 copy
  import math
  
  fin = None
  fout = None
  uabet = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V',
           'W', 'X', 'Y', 'Z']
  abet = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v',
          'w', 'x', 'y', 'z']
  INF = 10 ** 18
  
  
  def start(filename):
      filename = filename.replace('.py', '')
      global fin
      global fout
      infile = filename + ".in"
      outfile = filename + ".out"
      fin = open(infile)
      fout = open(outfile, "w")
  
  
  def permutation(lst):
      if len(lst) == 0:
          return []
      if len(lst) == 1:
          return [lst]
      l = []
      for i in range(len(lst)):
          m = lst[i]
          remLst = lst[:i] + lst[i + 1:]
          for p in permutation(remLst):
              l.append([m] + p)
      return l
  
  
  def subsets(x):
      retset = []
      if len(x) == 0:
          retset = [[]]
      else:
          first = x[-1]
          cop = x.copy()
          cop.remove(first)
          y = subsets(cop)
          retset.extend(copy.deepcopy(y))
          m = copy.deepcopy(y)
          for i in range(len(y)):
              k = m[0]
              m.remove(k)
              k.append(first)
              m.append(k)
          retset.extend(m.copy())
      return retset
  
  
  class Point:
      def __init__(self, x, y):
          self.x = x
          self.y = y
  
      def __str__(self):
          return str(self.x) + str(self.y)
  
      def __hash__(self):
          return hash(str(self))
  
      def __eq__(self, other):
          return self.x == other.x and self.y == other.y
  
  
  class frac:
      def __init__(self, n, d):
          self.n = n
          self.d = d
  
      def __str__(self):
          return str(self.n) + '/' + str(self.d)
  
      def __hash__(self):
          return hash(str(self))
  
      def simp(self):
          div = math.gcd(self.n, self.d)
          self.n = self.n / div
          self.d = self.d / div
  
      def __eq__(self, other):
          return self.n / self.d == other.n / other.d
  
      def __lt__(self, other):
          return self.n / self.d < other.n / other.d
  
  
  class Interval:
      def __init__(self, start, end):
          self.start = start
          self.end = end
  
      def __str__(self):
          return str(self.start) + str(self.end)
  
      def __hash__(self):
          return hash(str(self))
  
      def __eq__(self, other):
          return self.start == other.start and self.end == other.end
  
      def __lt__(self, other):
          return self.start < other.start
  
  
  def overlap(verl1, verl2):
      if verl1.end >= verl2.start:
          return True
      else:
          return False
  
  
  def merge(verl1, verl2):
      retverl = Interval(min(verl1.start, verl2.start), max(verl1.end, verl2.end))
      return retverl
  
  
  def changeBase(num, base):
      if base == 2:
          return "{0:b}".format(num)
      power = math.floor(math.log(num, base))
      remainder = num
      end = []
      iter = list(range(power + 1))
      iter.reverse()
      for i in iter:
          if remainder >= base ** i:
              q, r = divmod(remainder, base ** i)
              end.append(q)
              remainder = r
          else:
              end.append(0)
      retlist = []
      for i in end:
          if i > 9:
              retlist.append(uabet[i - 10])
          else:
              retlist.append(i)
      retlist = map(str, retlist)
      return ''.join(retlist)
  
  
  def prime(num):
      if num == 2:
          return True
      if num % 2 == 0:
          return False
      for i in range(2, math.ceil(num ** 0.5) + 1):
          if num % i == 0:
              return False
      return True
  
  
  def solve(s):
      z = s.count('0')
      o = s.count('1')
      if z==0:
          return 0
      if o == 0:
          return 1
      x = s.split('1')
      c= 0
      for  i in x:
          if i!='':
              c+=1
      if c<2:
          return c
      return 2
  
  def main():
      t = int(input())
      for i in range(t):
          s = input()
          print(solve(s))
  
  
  main()
  