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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.*;
  
  	public class practice{
  		public static void main(String[] args){
  			FastScanner r = new FastScanner();					
  			int t = r.i() ,n;
  			StringBuffer ans = new StringBuffer("");
  			char[] ar;
  			int min;
  			int i;
  			while(t-->0) {
  				ar =r.n().toCharArray();
  				min= ar[0]=='0' ? 1 : 0;
  				
  				for(i=1;i<ar.length;i++) {
  					
  					if(ar[i] != ar[i-1]) {
  						
  						min+=  ar[i]=='0' ? 1 : 0;
  					}
  				}
  				ans.append( Math.min(min, 2) + "\n");
  			}				
  			p(ans);
  		}
  
  		
  		static <T> void p(T s) {
  			System.out.println(s);
  		}
  		
  		static void p() {
  			System.out.println();
  		}		
  		static <T> void pn(T s) {
  			System.out.print(s);
  		}
  
  	}
  		
  
  	
  	class FastScanner {
  	    BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  	    StringTokenizer st=new StringTokenizer("");
  	    String n() {
  	        while (!st.hasMoreTokens())
  	            try {
  	                st=new StringTokenizer(br.readLine());
  	            } catch (IOException e) {
  	                e.printStackTrace();
  	            }
  	        return st.nextToken();
  	    }
  	    int[] ia(int n) {
  	    	int[] arr = new int[n];
  	    	for(int i=0;i<n;i++)
  	    		arr[i] = i();
  	    	return arr;
  	    }
  	    String[] sa(int n) {
  	    	String[] arr = new String[n];
  	    	for(int i=0;i<n;i++)
  	    		arr[i] = n();
  	    	return arr;
  	    }
  	    long[] la(int n) {
  	    	long[] arr = new long[n];
  	    	for(int i=0;i<n;i++)
  	    		arr[i] = l();
  	    	return arr;
  	    }
  	    String nl()
  	    {
  	        String str = "";
  	        try
  	        {
  	            str = br.readLine();
  	        }
  	        catch (IOException e)
  	        {
  	            e.printStackTrace();
  	        }
  	        return str;
  	    }
  
  	    int i() {
  	        return Integer.parseInt(n());
  	    }
  	    long l() {
  	        return Long.parseLong(n());
  	    }
  	}
  