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 {
  	public static void main(String args[]){
  		FScanner in = new FScanner();
  		PrintWriter out = new PrintWriter(System.out);
  		int t = in.nextInt();
  		while(t-->0) { 
  			String s=in.next();
  			int c0=0,c1=0,c2=0;
  
  			for(int i=0;i<s.length()-1;i++)
  			{
  				char ch=s.charAt(i);
  				int val=ch-'0';
  				if(ch=='0'&&(s.charAt(i+1)=='1'))
  				{
  					c0++;
  				}
  				
  				else if(ch=='1')
  					c1++;
  			}
  			if(s.charAt(s.length()-1)=='0')
  				c0++;
  
  			if(c0==0)
  				out.println(0);
  			else if(c0==1)
  				out.println(1);
  			else if(c0>1)
  				out.println(2);
  			
  		}
  		out.close();
  	}
  
  	static boolean checkprime(int n1)
  	{
  		if(n1%2==0||n1%3==0)
  			return false;
  		else 
  		{
  			for(int i=5;i*i<=n1;i+=6)
  			{
  				if(n1%i==0||n1%(i+2)==0)
  					return false;
  			}
  			return true;
  		}
  	}
  	
  	static class FScanner {
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer sb = new StringTokenizer("");
  
  		String next(){
  			while(!sb.hasMoreTokens()){
  				try{
  					sb = new StringTokenizer(br.readLine());
  				} catch(IOException e){ }
  			}
  			return sb.nextToken();
  		}
  		String nextLine(){
  			try{ return br.readLine(); } 
  			catch(IOException e) { } return "";
  		}
  
  		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] = nextInt();
  			return a;
  		}
  
  		float nextFloat(){
  		return Float.parseFloat(next());
  		}
  		
  		double nextDouble(){
  		return Double.parseDouble(next());
  		}
  	}
  }
  