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 java.util.*;
  import java.util.ArrayList;
  
  
  
  
  public class Main{
    
      public static void main(String[]args){
          Scanner sc=new Scanner(System.in);
          
      int t=sc.nextInt();
      for (int j=0;j<t;j++){
       String s=sc.next();
       int o=0;
       int l=0;
       for(int i=0;i<s.length();i++){
           if(s.charAt(i)-'0'==0)
           o++;
          
           
           else{
           l++;
               
           }
           
       }
       int oo=0;
       int lo=0;
   for(int i=0;i<s.length();i++){
        if(s.charAt(i)-'0'==0){
            oo++;
        while(i<s.length()&&s.charAt(i)-'0'==0){
        i++;}
        }
    if(i<s.length()&&s.charAt(i)-'0'==1){
        lo++;
          while(i<s.length()&&s.charAt(i)-'0'==1){
        i++;}
        
    }
    i--;
    
        
   }
   
       int ans=0;
       if(o==s.length())
       ans=1;
       if(l==s.length())
       ans=0;
       
       if(oo==1)
       ans=1;
       if(oo>=2)
       ans=2;
       System.out.println(ans);
  }
      }
  }