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: /*==========================================================================
  *   	AUTHOR:    RonWonWon
  *   	CREATED:   13.09.2021 17:50:07        
  /*==========================================================================*/
  import java.io.*;
  import java.util.*;
  
  public class B {
  	
  	public static void main(String[] args) throws IOException {
  		/*
      in.ini();
      PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("output.txt")));
      */
      long startTime = System.nanoTime();
      int t = in.nextInt(), T = 1;  
      while(t-->0) { 
        char ch[] = in.next().toCharArray();
        int c0 = 0, c1 = 0;
        int ans = 0, prev = 0;
        for(char c : ch){
        	if(c=='0'){
        		if(prev==1&&c0!=0){
        			ans = 2; break;
        		}
        		c0++; prev = 0;
        	}
        	else{
        		c1++;
        		prev = 1;
        	}
        }
        if(ans==0){
        	if(c0!=0) ans = 1;
        }
        out.println(ans);
     		//out.println("Case #"+T+": "+ans); T++;
      }
      /*
      startTime = System.nanoTime() - startTime;
      System.err.println("Time: "+(startTime/1000000));
      */
      out.flush();
  	}
  	
  	static FastScanner in = new FastScanner();
    static PrintWriter out = new PrintWriter(System.out);
  	static int oo = Integer.MAX_VALUE;
  	static long ooo = Long.MAX_VALUE;
  	
  	static class FastScanner {
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st = new StringTokenizer("");
  		
  		void ini() throws FileNotFoundException {
  			br = new BufferedReader(new FileReader("input.txt"));
  		}
  		
  		String next() {
  			while(!st.hasMoreTokens())
  				try { st = new StringTokenizer(br.readLine()); }
  				catch(IOException e) {}
  			return st.nextToken();
  		}
  		
  		String nextLine(){
  			try{ return br.readLine(); } 
  			catch(IOException e) { } return "";
  		}
  		
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		
  		int[] readArray(int n) {
  			int a[] = new int[n];
  			for(int i=0;i<n;i++) a[i] = nextInt();
  			return a;
  		}
  		
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  		
  		long[] readArrayL(int n) {
  			long a[] = new long[n];
  			for(int i=0;i<n;i++) a[i] = nextLong();
  			return a;
  		}
  		
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  		
  		double[] readArrayD(int n) {
  			double a[] = new double[n];
  			for(int i=0;i<n;i++) a[i] = nextDouble();
  			return a;
  		}
  	}
  
  	static final Random random = new Random();
  
  	static void ruffleSort(int[] a){
  		int n = a.length;
  		for(int i=0;i<n;i++){
  			int j = random.nextInt(n), temp = a[j];
  			a[j] = a[i]; a[i] = temp;
  		}
  		Arrays.sort(a); 	
  	}
  	
  	static void ruffleSortL(long[] a){
  		int n = a.length;
  		for(int i=0;i<n;i++){
  			int j = random.nextInt(n);
  			long temp = a[j];
  			a[j] = a[i]; a[i] = temp;
  		}
  		Arrays.sort(a); 	
  	}
  }
  