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: /* package codechef; // don't place package name! */
  
  import java.util.*;
  import java.lang.*;
  import java.io.*;
  
  /* Name of the class has to be "Main" only if the class is public. */
  public class Codechef
  {
  	public static void main (String[] args) throws java.lang.Exception
  	{
  		try {
  		   Scanner sc =new Scanner(System.in);
  		   int t=sc.nextInt();
  		   while(t-->0)
  		   {
  		       
  		       String s=sc.next();
  		       int zero=0;
  		       for(int i=0;i<s.length();i++)
  		       {
  		           if(s.charAt(i)=='0')
  		           zero++;
  		       }
  		       //find first postion of 0's
  		       int f_i=0,l_i=0;
  		       for(int i=0;i<s.length();i++)
  		       {
  		           if(s.charAt(i)=='0')
  		           {
  		               f_i=i; 
  		               break;
  		           }
  		       }
  		       //last index of 0's
  		       for(int i=s.length()-1;i>=0;i--)
  		       {
  		           if(s.charAt(i)=='0')
  		           {
  		               l_i=i; 
  		               break;
  		           }
  		       }
  		       //System.out.println(f_i+" "+l_i);
  		       if(zero==0)
  		       System.out.println(0);
  		       else if((l_i-f_i)+1==zero)
  		       System.out.println(1);
  		       else
  		       System.out.println(2);
  		   }
  		} catch(Exception e) {
  		} finally {
  		}
  	}
  }
  