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: // "static void main" must be defined in a public class.
  import java.util.*;
  public class Main {
      public static void main(String[] args) {
          Scanner  s=  new Scanner(System.in);
          int t=s.nextInt();
          for(int i=0;i<t;i++)
          {
             String str = s.next();
              
              System.out.println(fn(str));
          }
      }
      
      public static int fn(String str)
      {
      int nz=0;
          int no=0;
          int nt=0;
          int fz=-1;
          int lz=-1;
          int fo=-1;
          int lo=-1;
          int i=0;
          for(char c:str.toCharArray())
          {
              if(c=='0')
                 { nz++;  
                  
                
                  }
              else if(c=='1')
                 { no++; 
                  
                  }
              else
                  nt++;
              
              i++;
          }
          
          if(nz==0)
              return 0;
          
          if(no==0)
              return 1;
          
         int in=0;
          int tot=0;
          while(in<str.length())
          {
              char m  = str.charAt(in);
              in++;
              while(in<str.length() && str.charAt(in)==m)
                  m++;
             if(m=='0')
                 tot+=1;
          
          }
          
          return Math.min(2,tot);
          
          
      }
  }