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.io.BufferedReader;
  import java.io.FileReader;
  import java.io.IOException;
  import static java.lang.Math.*;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.Scanner;
  import java.util.StringTokenizer;
  import java.util.zip.ZipError;
  import java.io.InputStreamReader;
  import java.util.StringTokenizer;
  
  public class Solution {
  
      static int MOD = 998244353;
  
      static class FastReader {
          BufferedReader br;
          StringTokenizer st;
   
          public FastReader()
          {
              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 static void main(String[] args)throws IOException {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          FastReader s = new FastReader();
          int t = s.nextInt();
          for(int i = 0; i < t; i++ ){
              String str = s.nextLine();
              solve(str);
          }
      }
  
  
      static  void solve(String str){
          int n = str.length();
          int zeros  = 0;
          int ones = 0;
          boolean mid = false;
          int minzero = Integer.MAX_VALUE;
          int maxzero = Integer.MIN_VALUE;
          int res = 0;
          for(int i = 0; i < n; i++){
              if(str.charAt(i) == '0') {
                  zeros++;
                  minzero = Math.min(minzero, i);
                  maxzero = Math.max(maxzero, i);
              }
              else ones++;
          }
          for(int i = 0; i < n; i++){
              if(str.charAt(i) == '1') {
                  if(i < maxzero && i> minzero) mid = true;
              }
          }
          if(zeros == n) res = 1;
          else if(ones == n) res = 0;
          else{
              if(mid) res = 2;
              else res = 1;
          }
          System.out.println(res);
      }
  }