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: //package Div3.B;
  
  import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.ArrayList;
  import java.util.List;
  
  public class MINMEXCut {
      public static void main(String[] args) throws IOException {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          int t=Integer.parseInt(br.readLine());
          while(t>0){
              String str=br.readLine();
              List<Integer> sols=new ArrayList<>();
              sols.add(0);
              sols.add(1);
              sols.add(2);
              for(int i=0;i<str.length();i++){
                  int val=str.charAt(i)-'0';
                  if(sols.contains(val)){
                      sols.remove(Integer.valueOf(val));
                  }
              }
              int probableSolution = sols.get(0);
              int probableSolution2 = 0;
              for(int i=0;i<str.length();i++){
                  char c=str.charAt(i);
                  if(c=='0'){
                      while(i+1<str.length() && str.charAt(i+1)=='0')
                          i++;
                      probableSolution2+=1;
                  }
              }
              System.out.println(Math.min(probableSolution, probableSolution2));
              t--;
          }
      }
  }
  