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: 
  // Working program using Reader Class
  import java.io.DataInputStream;
  import java.io.FileInputStream;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.HashMap;
  import java.util.Scanner;
  import java.util.StringTokenizer;
  import java.lang.Math;
  
  public class Main {
  	public static void solve(String s) {
  
  		// reference link to submit your solution once system testing is done.
  		// link: https://codeforces.com/contest/1566/problem/B
  		int mex = 0;
  		int zeros = 0;
  		int ones = 0;
  		int n = s.length();
  		boolean two_substrings_present = false;
  
  		// count 0 and 1's occurances
  		for (int i = 0; i < n; i++) {
  			if (s.charAt(i) == '0')
  				zeros++;
  			if (s.charAt(i) == '1')
  				ones++;
  		}
  
  		// if all zeros, then answer is 1.
  		if (zeros == n)
  			mex = 1;
  
  		// if all ones, then answer is 0.
  		else if (ones == n)
  			mex = 0;
  
  		// if two substrings of 0, then answer is 2.
  		else {
  			// get first and last occurance of '0',
  			// and then check if '1' occurs between them.
  
  			int first = -1;
  			int last = -1;
  
  			// find first occurance
  			for (int i = 0; i < n; i++) {
  				if (s.charAt(i) == '0') {
  					first = i;
  					break;
  				}
  			}
  
  			// find last occurance
  			for (int i = n - 1; i > 0; i--) {
  				if (s.charAt(i) == '0') {
  					last = i;
  					break;
  				}
  			}
  
  			// check if '1' exist between first and last occurance of '0'
  			for (int i = first; i < last; i++) {
  				if (s.charAt(i) == '1') {
  					two_substrings_present = true;
  					mex = 2;
  					break;
  				}
  			}
  
  			if (!two_substrings_present)
  				mex = 1;
  		}
  
  		System.out.println(mex);
  		// System.out.println("n= " + n + " s= " + s + " median= " + median);
  	}
  
  	public static void main(String[] args) throws IOException {
  		// Reader sc = new Reader();
  		Scanner sc = new Scanner(System.in);
  		int tc = sc.nextInt();
  
  		while (tc-- > 0) {
  			String s = sc.next();
  			solve(s);
  		}
  		sc.close();
  	}
  
  }