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 Main {
  
    static class Pair {
  	  int fi;
  	  int se;
  	  
  	  public Pair(int fi, int se) {
  		  this.fi = fi;
  		  this.se = se;
  	  }
    }
  
    public static void main(final String[] args) {
      FastScanner sc = new FastScanner();
  
  	int t = sc.nextInt();
  	while(t-- > 0) {
  		String s = sc.next().replaceAll("1", " ").trim();
  		
  		if(s.length() > 0) {
  			System.out.println(Math.min(s.split("\\s+").length, 2));
  		} else {
  			System.out.println(0);
  		}
  	}
  
    } 
  
    static class FastScanner {
  		BufferedReader br;
  		StringTokenizer st;
  		public FastScanner() {
  			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;}
  	}
  }
  