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.*;
  
  public class Main {
  	static MyScanner sc = new MyScanner();
  	static Output out=new Output();
  	public static void main(String[] args) {
  		int t=sc.nextInt();
  		for(int i=0;i<t;i++) {
  			solve();
  		}
  		out.flush();
  	}
  	
  	public static void solve() {
  		String s=sc.next();
  		int c0=0,c1=0;
  		int l=s.length();
  		for(int i=0;i<l;i++) {
  			if(s.charAt(i)=='0')
  				c0++;
  			else
  				c1++;
  		}
  		if(c0==l)
  			out.println(1);
  		else if(c1==l)
  			out.println(0);
  		else {
  			int cc0=0;
  			for(int i=0;i<l;i++) {
  				if(s.charAt(i)=='0') {
  					for(int j=i;j<l;j++) {
  						if(s.charAt(j)=='0')
  							cc0++;
  						else
  							break;
  					}
  					break;
  				}
  			}
  			if(cc0==c0)
  				out.println(1);
  			else
  				out.println(2);
  		}
  	}	
  }
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  class MyScanner {
  
  	BufferedReader br;
  	StringTokenizer st;
  
  	public MyScanner() {
  		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 int[] nextIntArr(int n) {
  		int a[] = new int[n];
  		for (int i = 0; i < n; i++)
  			a[i] = nextInt();
  		return a;
  	}
  
  	public long[] nextLongArr(int n) {
  		long a[] = new long[n];
  		for (int i = 0; i < n; i++)
  			a[i] = nextLong();
  		return a;
  	}
  
  	public double[] nextDoubleArr(int n) {
  		double a[] = new double[n];
  		for (int i = 0; i < n; i++)
  			a[i] = nextDouble();
  		return a;
  	}
  
  	public String[] nextArr(int n) {
  		String a[] = new String[n];
  		for (int i = 0; i < n; i++)
  			a[i] = next();
  		return a;
  	}
  
  	public String[] nextLineArr(int n) {
  		String a[] = new String[n];
  		for (int i = 0; i < n; i++)
  			a[i] = nextLine();
  		return a;
  	}
  }
  class Output{
  	StringBuilder s=new StringBuilder();
  	public void println() {
  		s.append("\n");
  	}
  	public void println(Object x) {
  		s.append(x+"\n");
  	}
  	
  	public void print(Object x) {
  		s.append(x+"");
  	}
  	
  	public void flush() {
  		System.out.println(s);
  		s=new StringBuilder();
  	}
  }