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.*;
  import java.math.*;
  import java.lang.Math;
  
  public final class B_MIN_MEX_Cut {
      public static void main(String[] args) {
          MyScanner sc = new MyScanner();
          out = new PrintWriter(new BufferedOutputStream(System.out));
  
          // Start writing your solution here.
          int testCases = sc.nextInt();
          for (int cases = 0; cases < testCases; cases++) {
              // My code
              String binary = sc.next();
              boolean zero = binary.contains("0");
              boolean one = binary.contains("1");
  
              // checking if all zeros or all ones;
              if (!one || !zero) {
                  if (one) {
                      // all ones
                      out.println(0);
                  } else {
                      // all zeros
                      out.println(1);
                  }
  
                  // not all ones or zeros
              } else {
                  int firstIndex = binary.indexOf('0');
                  int lastIndex = binary.lastIndexOf('0');
                  int count = 0;
                  for (int i = 0; i < binary.length(); i++) {
                      if (binary.charAt(i) == '0') {
                          count++;
                      }
                  }
                  if (lastIndex - firstIndex + 1 != count) {
                      // there is a 1 in between
                      out.println(2);
                  } else {
                      out.println(1);
                  }
              }
          }
          // Stop writing your solution here.
          out.close();
      }
  
      // -----------PrintWriter for faster output---------------------------------
      public static PrintWriter out;
  
      // -----------MyScanner class for faster input----------
      public static class MyScanner {
          BufferedReader br;
          StringTokenizer st;
  
          public MyScanner() {
              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;
          }
      }
  }