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.util.*;
  
  public class Main {
      public static void main(String args[]) {
          FastReader fs = new FastReader();
          int t = fs.nextInt();
          while(t-- >0){
              String s = fs.next();
              int n = s.length();
              int cb = 0;
              int ans = 0;
              for(int i=0; i<n; i++){
                  if(s.charAt(i) == '0') cb = 1;
                  else if(s.charAt(i) == '1' && cb == 1){ans++; cb=0;}
              }
              if(s.charAt(n-1) == '0') ans++;
              System.out.println(Math.min(ans, 2));
          }
      }
  }
  
  class FastReader{
      BufferedReader br;
      StringTokenizer st;
  
      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());
      }
      int[] nextArr(int n){
          int[] a = new int[n];
          for(int i=0; i<n; i++){
              a[i] = nextInt();
          }
          return a;
      }
  }