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.BufferedWriter;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.OutputStreamWriter;
  import java.util.Scanner;
  
  public class Codeforcesmaytwentyone {
      public static void main(String[] args) throws IOException {
          try {
              BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
              BufferedWriter log = new BufferedWriter(new OutputStreamWriter(System.out));
              
               int t = Integer.parseInt(br.readLine().trim());
               while(t-->0){
                  String s = br.readLine(); 
                  char arr[] = s.toCharArray();
                  long sum =0;
                    long c=0;
                    for(int i=0; i<s.length(); i++){
                        if(arr[i]=='0')c=1;
                        else if((arr[i]!=0 &&c==1)){
                            sum++; c=0;
                        }
                    }
                 if(arr[s.length()-1]=='0')sum++;
                 System.out.println(Math.min(sum, 2));
                    
               }
              br.close();
              log.flush();
          } catch(Exception e) {
              return;
          }
      }   // end-main
  
      static void print(int arr[]) {
          System.out.print("Print :- ");
          for(int i=0 ; i<arr.length ; i++) {
              System.out.print(arr[i]+" ");
          }
          System.out.println();
      }   // end-print
      static boolean isPrime(int n)
      {
   
          // Check if number is less than
          // equal to 1
          if (n <= 1)
              return false;
   
          // Check if number is 2
          else if (n == 2)
              return true;
   
          // Check if n is a multiple of 2
          else if (n % 2 == 0)
              return false;
   
          // If not, then just check the odds
          for (int i = 3; i <= Math.sqrt(n); i += 2)
          {
              if (n % i == 0)
                  return false;
          }
          return true;
      }
   
   
  }
  