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.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.StringTokenizer;
  
  
  public class B {
  	public static void main(String[] args)
      {
          FastReader s=new FastReader();
          int t = s.nextInt();         // Number of test cases
          for (int i = 0; i<t; i++) {
          	String str = s.next();
          	boolean allz = true;
          	boolean allo = true;
          	boolean flag = false;
          	int countzoz = 0;
          	for (int j = 0; j<str.length(); j++) {
          		char ch = str.charAt(j);
          		if (ch == '0' && countzoz == 0) {
          			countzoz++;
          			allo = false;
          		}else if (ch == '0' && countzoz == 2) {
          			flag = true;
          			System.out.println(2);                              // We found 010 formal, meaning answer is 2
          			break;
          		}
          		if (ch == '1' && countzoz == 1) {
          			countzoz++;
          			allz = false;
          		}
          	}
          	if (allo == true) System.out.println(0);
          	else if (allz == true) System.out.println(1);
          	else if (!flag) System.out.println(1);
  
          }
  
      
          
          
      }
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
      
  	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;
          }
      }
  	
  }
  