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.*;
  import java.lang.*;
  
  public class C2Q2 {
      public static void main(String[] args) throws IOException {
          BufferedReader br = new BufferedReader( new InputStreamReader(System.in));
          
          int t=Integer.parseInt(br.readLine());
          
          BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out));
          C2Q2 obj1=new C2Q2();
          while(t-->0){
              String s;
              s=br.readLine();
              output.write(obj1.solve(s)+"\n");
          }
          output.flush();
      }
  
      public int solve(String s){
          int k=0;
          int zgrps=0;
          int n=s.length();
              while(k<n){
                  if(s.charAt(k)=='1'){
                      k++;
  
                  }else{
                      zgrps++;
                      while( (k<n)&&(s.charAt(k)=='0') ){
                          k++;
                      } 
                  }
              }
              return  Math.min(2,zgrps);
          }
  
  
      }
  
  