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 B1566{
  	static Scanner sc = null;
  	public static void main(String args[]){
  		sc = new Scanner(System.in);
  		int t = sc.nextInt();
  		sc.nextLine();
  		while(t-->0){
  			String s = sc.nextLine();
  			int N = s.length();
  			ArrayList<Integer> ar = new ArrayList<>();
  			char ch = s.charAt(0);
  			ar.add(Integer.parseInt(Character.toString(ch)));
  			for(int i=1;i<N;i++){
  				char c1 = s.charAt(i);
  				if(ch!=c1){
  					ar.add(Integer.parseInt(Character.toString(c1)));
  				}
  				ch = c1;	
  			}
  			int size = ar.size();
  			int count0 = 0;
  			int count1 = 0;
  			for(Integer j : ar){
  				if(j==0)
  					count0+=1;
  				else
  					count1+=1;
  			}
  			int ans = 0;
  			if(size>=4){
  				ans = 2;
  			}
  			else if(size==3){
  				if(count1==2)
  					ans = 1;
  				else
  					ans = 2;
  			}
  			else if(size==2){
  				ans = 1;
  			}
  			else{
  				if(count1==1)
  					ans = 0;
  				else
  					ans = 1;
  			}
  			System.out.println(ans);
  		}
  	}
  }
  
  