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.ArrayList;
  import java.util.Arrays;
  import java.util.HashMap;
  import java.util.HashSet;
  import java.util.Map;
  import java.util.Random;
  import java.util.Set;
  import java.util.StringTokenizer;
  import java.util.TreeMap;
  import java.util.TreeSet;
  import java.io.*;
  
  	public class A {
  		
  		BufferedReader br;
  		StringTokenizer st;
  		
  		
  		public A(){ // constructor
  			
  			br = new BufferedReader(new InputStreamReader(System.in));
  		}
  		
  		String next(){
  			
  			while (st == null || !st.hasMoreElements()){
  				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());
  		}
  		
  		double nextDouble(){
  			return Double.parseDouble(next());
  		}
  
  		String nextLine(){
  			
  			String str = "";
  			try{
  				
  				str = br.readLine();
  			}
  			catch (IOException e){
  				
  				e.printStackTrace();
  			}
  			return str;
  		}
  		
  		static void sort(int[] arr){
  			
  	        int n = arr.length;
  	        
  	        Random rnd = new Random();
  	        
  	        for(int i = 0; i < n; ++i){
  	        	
  	            int tmp = arr[i];
  	            int randomPos = i + rnd.nextInt(n - i);
  	            arr[i] = arr[randomPos];
  	            arr[randomPos] = tmp;
  	        }
  	        
  	        Arrays.sort(arr);
  	   }
  		
  		static class Pair{
  			int loc;
  			int speed;
  			Pair(int loc, int speed){
  				this.loc = loc;
  				this.speed = speed;
  			}
  		}
  		
  		static long gcd_long(long a, long b) {
  			if(b == 0) {
  				return a;
  			}
  			return gcd_long(b, a % b);
  		}
  		
  		static int mod = (int)1e9 + 7;
  		static long mul_long = 1;
  		
  	public static void main(String[] args) throws Exception{
  		
  		A in = new A();
  		PrintWriter out = new PrintWriter(System.out);
  		
  		// hard work will power dedication
  		
  		// Rippling CNIL Spri GOOG.
  		
  		int test = in.nextInt();
  		while(test-- > 0) {
  			
  			String str = in.next();
  			int ans = 0;
  			
  			for(int i = 0; i < str.length(); i++) {
  				char ch = str.charAt(i);
  				
  				if(ch == '0'){
  					ans++;
  					while(i < str.length() && str.charAt(i) == '0') {
  						i++;
  					}
  					
  				}
  			}
  			out.println(Math.min(2, ans));
  		}
  		
  		out.flush();
  		out.close();
  	}
  }