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.util.*;
  import java.lang.*;
  import java.io.*;
   
  public class main
  {
      
      
      
  	public static void main (String[] args) throws java.lang.Exception
      {
  		FastReader sc=new FastReader();
              int t = sc.nextInt();
              while(t-->0){
                  String s = sc.next();
                  
                  int n = s.length();
                  int i = 0;
                  int mex = 0;
                  while(i<n){
  if(s.charAt(i) == '0' ){
  while(i<n && s.charAt(i)=='0'){
                          i++;
                         
                      }
  mex++;
  
  }
                      
                      while(i<n && s.charAt(i) == '1'){
                          i++;
                      }
                  }
                  
                  System.out.println(Math.min(2,mex));
              }  
      }
  }
  
  class pair{
      int v;
      int i;
      pair(int v , int i){
          this.v = v;
          this.i = i;
      }
  }
   
  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;}
  }