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 MIN_MEX_Cut {
  
  	static 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;
  		}
  
  	}
  
  	public static void shuffle(int[] a) {
  		Random r = new Random();
  
  		for (int i = 0; i <= a.length - 2; i++) {
  			int j = i + r.nextInt(a.length - i);
  
  			swap(a, i, j);
  		}
  
  		Arrays.sort(a);
  	}
  
  	public static void swap(int[] a, int i, int j) {
  		int temp = a[i];
  		a[i] = a[j];
  		a[j] = temp;
  	}
  
  	public static void main(String[] args) {
  		// TODO Auto-generated method stub
  		FastReader t = new FastReader();
  		PrintWriter o = new PrintWriter(System.out);
  		int test = t.nextInt();
  
  		while (test-- > 0) {
  			char[] ch = t.next().toCharArray();
  			int count = ch[0] == '0' ? 1 : 0;
  
  			for (int i = 0; i < ch.length - 1; ++i)
  				if (ch[i] == '1' && ch[i + 1] == '0')
  					++count;
  
  			o.println(count == 0 ? 0 : count == 1 ? 1 : 2);
  		}
  
  		o.flush();
  		o.close();
  	}
  }