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.io.*;
  public class Main {
  // For fast input output
  static class FastReader {
  BufferedReader br;
  StringTokenizer st;
  public FastReader()
  { try {br = new BufferedReader(
  new FileReader("input.txt"));
  PrintStream out = new PrintStream(new FileOutputStream("output.txt"));
  System.setOut(out);}
  catch(Exception e) { 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;
  }
  }
  // end of fast i/o code
  public static void main(String[] args) {
    FastReader reader = new FastReader();
    int Q = reader.nextInt();
    outer: for (int q = 0; q < Q; q++) {
     String a = reader.nextLine();
     int N = a.length();
     int zeros = 0, ones = 0;
     for (int i = 0; i < N - 1; i++) {
       if (a.charAt(i) != a.charAt(i + 1)) {
         if (a.charAt(i) == '0') zeros++;
         else ones++;
       }
      }
      if (a.charAt(N - 1) == '0') zeros++;
      else ones++;
      System.out.println(Math.min(2, zeros));
     
    }
  }}