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.io.IOException;
  import java.io.InputStreamReader;
  import java.math.BigInteger;
  import java.util.*;
  public class b729 {
  
  	public static void main(String[] args) throws IOException {
  			FastReader sc = new FastReader();
  	
  	int t = sc.nextInt();
  	
  for(int o = 0 ; o<t;o++) {
  
  	String s = sc.next();
  	int n = s.length();
  	int c = 0;
  	for(int i = 0 ; i<n;i++) {
  		if(s.charAt(i) == '1') {
  			c++;
  		}
  	}
  	int d = n-c;
  	if(c==0) {
  		System.out.println(1);
  	}else if(c==n) {
  		System.out.println(0);
  	}else if(c == 1) {
  		if(s.charAt(0)=='1' || s.charAt(n-1)=='1') {
  			System.out.println(1);
  		}else {
  			System.out.println(2);
  		}
  	}else if(d==1) {
  		System.out.println(1);
  	}
  	else {
  		int f = 0;
  		for(int i = 0 ; i<n;i++) {
  			if(s.charAt(i) =='0' ) {
  				int j = i;
  				int x= 0;
  				while(j<n && s.charAt(j) == '0') {
  					x++;
  					j++;
  				}
  			if(x == d) {
  				f = 1;
  			}
  			break;
  			}
  			
  		}
  		
  		if(f == 1) {
  			System.out.println(1);
  		}
  		else
  		System.out.println(2);
  	}
  	
  }
  	
  	}
  }
  
  
  class FastReader {
      BufferedReader br;
      StringTokenizer st;
  
      public FastReader()
      {
          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;
      }
  }