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 math,sys
  #from itertools import permutations, combinations;import heapq,random;
  from collections import defaultdict,deque
  import bisect as bi
  def yes():print('YES')
  def no():print('NO')
  #sys.stdin=open('input.txt','r');sys.stdout=open('output.txt','w');
  def I():return (int(sys.stdin.readline()))
  def In():return(map(int,sys.stdin.readline().split()))
  def Sn():return sys.stdin.readline().strip()
  def ceil(a,b):return ((a+b-1)//(b))
  #sys.setrecursionlimit(1500)
  def dict(a):
      d={} 
      for x in a:
          if d.get(x,-1)!=-1:
              d[x]+=1
          else:
              d[x]=1
      return d
  def find_gt(a, x):
      'Find leftmost value greater than x'
      i = bi.bisect_left(a, x)
      if i != len(a):
          return i
      else:            
          return -1
  
  
  def main():
      try:
  
          s=Sn()
          ap=[]
          cnt=0
          last=s[0]
          for x in s[1:]:
              if x!=last:
                  ap.append(last)
                  last=x
          ap.append(last)
          cnt=0
          for x in ap:
              if x=='0':
                  cnt+=1
          print(min(cnt,2))
  
      except:
          pass
          
  M = 998244353
  P = 1000000007
   
  if __name__ == '__main__':
      for _ in range(I()):main()
      # for _ in range(1):main()
  #End#
  
  
  #        ******************* All The Best *******************   #