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 B {
  
      static int mod = 1000000007;
  
      public static void main(String[] args) throws IOException {
  
          // File file = new File("input.txt");
          // BufferedReader br = new BufferedReader(new FileReader(file));
  
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          BufferedWriter log = new BufferedWriter(new OutputStreamWriter(System.out));
  
          boolean isIt = true;
          int t = Integer.parseInt(br.readLine().trim()), ans = 0, n = 0;
          while (t-- > 0) {
              char arr[] = br.readLine().toCharArray();
              n = arr.length;
              ans = 0;
              isIt = true;
  
              for (int i = 0; i < n; i++) {
                  if (isIt) {
                      if (arr[i] == '0') {
                          ans++;
                          isIt = false;
                      }
                  }
  
                  if (arr[i] == '1') {
                      isIt = true;
                  }
  
                  if (ans > 2) {
                      ans = 2;
                      break;
                  }
              }
  
              log.write(ans + "\n");
          }
  
          br.close();
          log.flush();
      } // end-main
  
  } // end-class
  