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 static java.lang.Integer.parseInt;
  import static java.lang.Long.parseLong;
  import static java.lang.System.exit;
  
  import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.util.StringTokenizer;
  
  public class B {
  
  	static void solve() throws Exception {
  		int tests = scanInt();
  		test: for (int test = 0; test < tests; test++) {
  			String s = scanString();
  			int i, j;
  			for (i = 0; i < s.length() && s.charAt(i) == '1'; i++) { }
  			if (i == s.length()) {
  				out.println(0);
  				continue;
  			}
  			for (j = s.length(); j > 0 && s.charAt(j - 1) == '1'; j--) { }
  			for (int k = i; k < j; k++) {
  				if (s.charAt(k) != '0') {
  					out.println(2);
  					continue test;
  				}
  			}
  			out.println(1);
  		}
  	}
  
  	static int scanInt() throws IOException {
  		return parseInt(scanString());
  	}
  
  	static long scanLong() throws IOException {
  		return parseLong(scanString());
  	}
  
  	static String scanString() throws IOException {
  		while (tok == null || !tok.hasMoreTokens()) {
  			tok = new StringTokenizer(in.readLine());
  		}
  		return tok.nextToken();
  	}
  
  	static BufferedReader in;
  	static PrintWriter out;
  	static StringTokenizer tok;
  
  	public static void main(String[] args) {
  		try {
  			in = new BufferedReader(new InputStreamReader(System.in));
  			out = new PrintWriter(System.out);
  			solve();
  			in.close();
  			out.close();
  		} catch (Throwable e) {
  			e.printStackTrace();
  			exit(1);
  		}
  	}
  }