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.lang.*;
  import java.io.*;
  
  public class B_MIN_MEX_Cut
  {
  	public static void main (String[] args) throws java.lang.Exception
  	{
  		// your code goes here
  		
  		FastReader s = new FastReader();
  		
  		int t = s.nextInt();
  		
  		while(t-- != 0){
  		    String st = s.nextLine();
  
              int noss = 0;
  
              for(int i = 1;i<st.length();i++){
                  if(st.charAt(i-1) == '1' && st.charAt(i) == '0'){
                      noss++;
                  }
              }
  
              if(st.charAt(0) == '0'){
                  noss++;
              }
  
              if(noss>1){
                  System.out.println(2);
              }
              else{
                  System.out.println(noss);
              }
  			
  		}
  	}
  	
  	public static int ans(long l, long r, long v){
  		
  //		System.out.println((r-l)/2);
  		
  		long m = (r+l)/2;
  		
  //		System.out.println(m);
  		
  		if(((46*(m*m))+(6*m)+4)>=v){
  			return 1;
  		}
  		else{
  			return 0;
  		}
  	}
  	
  	static class FastReader 
      { 
          BufferedReader br; 
          StringTokenizer st; 
    
          public FastReader() 
          { 
              br = new BufferedReader(new
                       InputStreamReader(System.in)); 
          } 
    
          String next() 
          { 
              while (st == null || !st.hasMoreElements()) 
              { 
                  try
                  { 
                      st = new StringTokenizer(br.readLine()); 
                  } 
                  catch (IOException  e) 
                  { 
                      e.printStackTrace(); 
                  } 
              } 
              return st.nextToken(); 
          } 
    
          int nextInt() 
          { 
              return Integer.parseInt(next()); 
          } 
    
          long nextLong() 
          { 
              return Long.parseLong(next()); 
          } 
    
          double nextDouble() 
          { 
              return Double.parseDouble(next()); 
          } 
    
          String nextLine() 
          { 
              String str = ""; 
              try
              { 
                  str = br.readLine(); 
              } 
              catch (IOException e) 
              { 
                  e.printStackTrace(); 
              } 
              return str; 
          } 
      } 
  }
  
  