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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.*;
  public class B{
  	public static void main(String[] args){
  		FastReader sc = new FastReader();
  		int t=sc.nextInt();
  		while(t-->0){
  			char c[]=sc.next().toCharArray();
  			int count0=0;
  			for(int i=0;i<c.length;i++) {
  				if(c[i]=='0')count0++;
  			}
  			if(count0==0) {
  				System.out.println(0);
  			}else {
  				int check=0;
  				for(int i=0;i<c.length;i++) {
  					if(c[i]=='0'|| check!=0) {
  						if(c[i]!='0')break;
  						check++;
  					}
  				}
  				if(check==count0) {
  					System.out.println(1);
  				}else System.out.println(2);
  			}
  			
  		}		 
      }
  
  	static class pair {
  		int x;int y;
  		 pair(int x,int y){
  			this.x=x;
  			this.y=y;
  		}
  		
  	}
  	static ArrayList<Long> prime_in_given_range(long l,long r){
  		ArrayList<Long> ans= new ArrayList<>();
  		int n=(int)Math.sqrt(r)+1;
  		int prime[]=sieve_of_Eratosthenes(n);
  		long res[]=new long [(int)(r-l)+1];
  		for(int i=0;i<=r-l;i++) {
  			res[i]=i+l;
  		}
  		for(int i=0;i<prime.length;i++) {
  			if(prime[i]==1) {
  				System.out.println(2);
  				for(int j=Math.max((int)i*i, (int)(l+i-1)/i*i);j<=r;j+=i) {
  					res[j-(int)l]=0; 
  				}
  			}
  		}
  		for(long i:res) if(i!=0)ans.add(i);
  		return ans;
  	}
  	static int [] sieve_of_Eratosthenes(int n) {
  		int prime[]=new int [n];
  		Arrays.fill(prime, 1);  // 1-prime | 0-not prime
  		prime[0]=prime[1]=0;
  		for(int i=2;i<n;i++) {
  			if(prime[i]==1) {
  				for(int j=i*i;j<n;j+=i) {
  					prime[j]=0;
  				}
  			}
  		}
  		return prime;
  	}
  	static long binpow(long a,long b) {
  		long res=1;
  		if(b==0)return a;
  		if(a==0)return 1;
  		while(b>0) {
  			if((b&1)==1) {
  				res*=a;
  			}
  			a*=a;
  			b>>=1;
  		}
  		return res;
  	}
  	static void print(int a[]) {
  		System.out.println(a.length);
  		for(int i:a) {
  			System.out.print(i+" ");
  		}
  		System.out.println();
  	}
  	static void print(long a[]) {
  		System.out.println(a.length);
  		for(long i:a) {
  			System.out.print(i+" ");
  		}
  		System.out.println();
  	}
  	static void print(ArrayList<Long> a) {
  		System.out.println(a.size());
  		for(long i:a) {
  			System.out.print(i+" ");
  		}
  		System.out.println();
  	}
  	static class FastReader {
  		BufferedReader br;
  		StringTokenizer st;
  
  		public FastReader()
  		{
  			br = new BufferedReader(
  				new InputStreamReader(System.in));
  		}
  		int [] fastArray(int n) {
  			int a[]=new int [n];
  			for(int i=0;i<n;i++) {
  				a[i]=nextInt();
  			}
  			return a;
  		}
  		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;
  		}
  	}
  
  }