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: //package Codeforces;
  import java.util.*;
  import java.io.*;
  
  public class template{
  	static class FastScanner {
  		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st = new StringTokenizer("");
  		String next () {
  			while(!st.hasMoreTokens()) {
  				try {
  					st=new StringTokenizer(br.readLine());
  				}catch(IOException e) {
  					e.printStackTrace();
  				}
  			}
  			return st.nextToken();
  		}
  		
  		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());
  		}
  	}
  	public static int gcd(int a,int b) {
  		if(b == 0) {
  			return a;
  		}
  		return gcd(b,a%b);
  	}
  	
  	public static void main(String[] args) {
  		 FastScanner sc = new FastScanner();
  		 int test = sc.nextInt();
  		 while(test-->0) {
  			 String s = sc.next();
  			 String onestr = s.replace("1","");
  			 String zerostr = s.replace("0","");
  			 if(zerostr.length() == 0) {
  				 System.out.println("1");
  				 continue;
  			 }
  			 if(onestr.length() == 0) {
  				 System.out.println("0");
  				 continue;
  			 }
  			 int count = 0;
  			 int c = 0;
  			 int i;
  			 for( i=0;i<s.length()-1;i++) {
  				 if((s.charAt(i) == '0' && s.charAt(i+1) == '1') ){
  					 count++;
  				 }
  				 if((s.charAt(i) == '1' && s.charAt(i+1) == '0')) {
  					 c++;
  				 }
  				
  			 }
  			 if(count == 1 && c == 1 && s.charAt(0) == '1' && s.charAt(s.length()-1) == '1') {
  				 System.out.println("1");
  			 }
  			 else if(count+c == 1) {
  				 System.out.println("1");
  			 }
  			 else {
  				 System.out.println("2");
  			 }
  		 }
        
  	}
  }
  //6
  //01
  //1111
  //01100
  //101
  //0000
  //01010
  //1
  //0
  //2
  //1
  //1
  //2
  class pair {
  	int value;
  	int index;
  	pair(int value,int index){
  		this.value = value;
  		this.index = index;
  	}
  	public int getValue() {
  		return value;
  	}
  	public int getIndex() {
  		return index;
  	}
  	public void setIndex(int index) {
  		this.index = index;
  	}
  	public void setValue(int value) {
  		this.value = value;
  	}
  }
  
  