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.*;
  import java.io.*;
  public class B{
  	public static void main(String[] args)
  	{
  		FastScanner fs = new FastScanner();
  		PrintWriter out = new PrintWriter(System.out);
  		int t = fs.nextInt();
  		for(int tt=0;tt<t;tt++){
  		char[] arr = fs.next().toCharArray();
  		int min = 2;
  		
  		int one = 0; int zero = 0;
  		
  		for(char ch:arr)
  		{
  		    if(ch == '1')one++;
  		    else zero++;
  		}
  		
  		if(zero == 0)
  		{
  		    min = 0;
  		}
  		else if(one == 0)
  		{
  		    min = 1;
  		}
  		else{
  		int count = 0;
  		
  		if(arr[0] == '0')count++;
  		
  		for(int i=1;i<arr.length;i++)
  		{
  		    if(arr[i] == '0' && arr[i-1] != '0')count++;
  		}
  		
  		if(count<2)min = 1;
  		}
  		
  		out.println(min);
  		}
  		
  		out.close();		
  	}
  	static class FastScanner {
  		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st=new StringTokenizer("");
  		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());
  		}
  		int[] readArray(int n) {
  			int[] a=new int[n];
  			for (int i=0; i<n; i++) a[i]=nextInt();
  			return a;
  		}
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  	}
  	public static int[] sort(int[] arr)
  	{
  		List<Integer> temp = new ArrayList();
  		for(int i:arr)temp.add(i);
  		Collections.sort(temp);
  		int start = 0;
  		for(int i:temp)arr[start++]=i;
  		return arr;
  	}
  	public static String rev(String str)
  	{
  		char[] arr = str.toCharArray();
  		char[] ret = new char[arr.length];
  		int start = arr.length-1;
  		for(char i:arr)ret[start--] = i;
  		String ret1 = "";
  		for(char ch:ret)ret1+=ch;
  		return ret1;
  	}
  }