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: from sys import stdin,stdout
  def INPUT():return list(int(i) for i in stdin.readline().split())
  def inp():return stdin.readline()
  def out(x):return stdout.write(x)
  from math import *
  from decimal import *
  INT_MAX=10**13
  # from bisect import bisect_left
  # import sys
  # MOD=10**9+7
  # from random import randint as R
  # from heapq import heappush,heappop,heapify
  #320660CC
  mod=998244353
  #=============================================
  def ceil(a,b):
      return (a+b-1)//b
  def pow(a,n):
      res=1
      while(n):
          if n&1:
              res=(res*a)%mod
          a=a*a
          a%=mod
          n>>=1
      return res
  for _ in range(int(input())):
      s=input()
      n=len(s)
      i=0
      ans=0
      if '1' in s and '0' in s:
          fin=2
      else:
          fin=mod**2
      while(i<n):
          if s[i]=='0':
              ans+=1
              while(i<n and s[i]=='0'):
                  i+=1
          else:
              i+=1
      print(min(ans,fin))
  