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.Scanner;
  
  public class Main {
  
      public static void main(String[] args) {
          Scanner sc = new Scanner(System.in);
          int t = sc.nextInt();
          sc.nextLine();
  
  
          while (t-- != 0) {
              String s = sc.nextLine();
  //            int num = Integer.parseInt(s);
              int l = s.length();
             int zeroes = 0;
             for(int i=0; i<l; i++)
             {
                 char ch = s.charAt(i);
                 if(ch=='0')
                     zeroes++;
             }
             if(zeroes==0)
             {
                 System.out.println(0);
                 continue;
             }
  //           boolean printed = false;
             int start = 0, end = l-1;
             while(start<=end)
             {
                 if(s.charAt(start)=='0' && s.charAt(end)=='0')
                 {
                     if(end-start+1==zeroes)
                     {
  
                         System.out.println(1);
                     }
                     else
                     {
                         System.out.println(2);
                     }
                     break;
                 }
                 if(s.charAt(start)!='0')
                     start++;
                 if(s.charAt(end)!='0')
                     end--;
  
             }
  
  
          }
      }
  
  
  }
  