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.util.*;
  import java.text.*;
  import java.math.*;
  import java.util.regex.*;
  
  public class JaiShreeRam{
  	static Scanner in=new Scanner();
  	static long mod = 1000000007;
  	static ArrayList<ArrayList<Integer>> adj;
  	public static void main(String[] args) throws Exception{
  		int z=in.readInt();
  		while(z-->0){
  			solve();
  		}
  	}
  		
  	static void solve() {
  		char c[]=in.readString().toCharArray();
  		int n=c.length;
  		Set<Character> st=new HashSet<>();
  		for(int i=0;i<n;i++) {
  			st.add(c[i]);
  		}
  		if(!st.contains('0')) {
  			System.out.println(0);
  		}
  		else if(!st.contains('1')){
  			System.out.println(1);
  		}
  		else {
  			int ans=2;
  			int cnt=0;
  			int k=0;
  			for(int i=0;i<n;i++) {
  				if(c[i]=='0') {
  					cnt++;
  				}
  				else {
  					if(cnt>0) {
  						k++;
  						cnt=0;
  					}
  				}
  			}
  			if(cnt>0) {
  				k++;
  				cnt=0;
  			}
  			System.out.println(Math.min(k, ans));
  		}
  	}
  	
  	static int[] nextIntArray(int n){
  		int[] arr= new int[n];
  		int i=0;
  		while(i<n){
  			arr[i++]=in.readInt();
  		}
  		return arr;
  	}
  	static int[] nextIntArray1(int n){
  		int[] arr= new int[n+1];
  		int i=1;
  		while(i<=n){
  			arr[i++]=in.readInt();
  		}
  		return arr;
  	}
  	static long gcd(long a, long b) {
  		if (b==0) return a;
  		return gcd(b, a%b);
  	}
  	static class Scanner{
  		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st=new StringTokenizer("");
  		String readString() {
  			while (!st.hasMoreTokens())
  				try {
  					st=new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			return st.nextToken();
  		}
  		double readDouble() {
  			return Double.parseDouble(readString());
  		}
  		int readInt() {
  			return Integer.parseInt(readString());
  		}
  		long readLong() {
  			return Long.parseLong(readString());
  		}
  	}	
  }