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.*;
  /*
  
  */
  
   
   public class B{
  	static FastReader sc=null;
  	
  	public static void main(String[] args) {
  		sc=new FastReader();
  		int t=sc.nextInt();
  		while(t-->0) {
  			char a[]=sc.next().toCharArray();
  			int n=a.length;
  			int cnt=1;
  			for(int i=0;i+1<n;i++)if(a[i]!=a[i+1])cnt++;
  			
  			if(a[0]=='0')cnt++;
  			cnt/=2;
  			int ans=Math.min(2,cnt);
  			System.out.println(ans);
  		}
  	}
  	
  	static int[] ruffleSort(int a[]) {
  		ArrayList<Integer> al=new ArrayList<>();
  		for(int i:a)al.add(i);
  		Collections.sort(al);
  		for(int i=0;i<a.length;i++)a[i]=al.get(i);
  		return a;
  	}
  	
  	static void print(int a[]) {
  		for(int e:a) {
  			System.out.print(e+" ");
  		}
  		System.out.println();
  	}
  	
  	static class FastReader{
  		
  		StringTokenizer st=new StringTokenizer("");
  		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  		
  		String next() {
  			while(!st.hasMoreTokens()) 
  				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());
  		}
  		
  		int[] readArray(int n) {
  			int a[]=new int[n];
  			for(int i=0;i<n;i++)a[i]=sc.nextInt();
  			return a;
  		}
  	}
  	
  	
  }
  