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.*;
  public class MyClass {
      
      public static int getLength(String str)
      {
          int n=str.length();
          if(n>0)
          return n;
          else
          return 0;
      }
      public static void reddyAns(int a,int b)
      {
          if(a<=2)
            System.out.println(a);
          else if(a<0)
            System.out.println("-1");
          else if(b==0)
            System.out.println("1");
          else if(a==0)
            System.out.println("0");
          else if(b<0)
            System.out.println("-8");
          else
            System.out.println("2");
      }
      
      public static boolean checkFRE(int k)
      {
          if(k>0)
          return true;
          else
          return false;
      }
      
      public static int hadi(int z)
      {
          int e=z+1;
          return e;
      }
      
      public static int buzzi(int l)
      {
          int r=l+1;
          return r;
      }
      
      public static void main(String args[]) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        int x,y,z,kii;
        String str;
        for(int i=0;i<t;i++)
        {
            str=sc.next();
            int yeshw=9;
            int reddy=7;
            z=yeshw-reddy;
            z=2;
            x=0;
            y=0;
            int kit=0;
            int n=getLength(str);
            for(int yesh=0;yesh<n;yesh++)
            {
                char ch=str.charAt(yesh);
                if(ch!=z)
                {
                    if(ch=='1'){
                      y=hadi(y);
                    }
                    else if(ch!='0' && ch!='1')
                    kit++;
                    else{
                        x=buzzi(x);
                    }
                    z=ch;
                    kit++;
                    boolean fi=checkFRE(kit);
                    if(!fi)
                    {
                        break;
                    }
                }
                for(int o=0;o<1;o++)
                {
                    break;
                }
                
            }
            reddyAns(x,y);
            
        }
      }
  }