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: // Piyush Nagpal
  import java.util.*;
  import java.io.*;
   public class C{
       
      static int MOD=1000000007;
      static PrintWriter pw;
      static FastReader sc;
      
      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());}
          
          public char nextChar() throws IOException {return next().charAt(0);}
   
          String nextLine(){
              String str = "";
              try {
                  str = br.readLine();
              }
              catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
          public int[] intArr(int a,int b,int n) throws IOException {int[]in=new int[n];for(int i=a;i<b;i++)in[i]=nextInt();return in;}
          public long[] longArr(int a,int b,int n) throws IOException {long[]in=new long[n];for(int i=a;i<b;i++)in[i]=nextLong();return in;}
          
      }   
   
   
       public static long gcd(long a,long b){if( b>a ){return gcd(b,a);}if( b==0 ){return a;} return gcd(b,a%b);}
       public static long expo( long a,long b,long M ){long result=1;while(b>0){if( b%2==1 ){result=(result*a)%M;}a=(a*a)%M;b=b/2;}return result%M;}
      public static ArrayList<Long> Sieve(int n){boolean [] arr= new boolean[n+1];Arrays.fill(arr, true);ArrayList<Long> list= new ArrayList<>();for(int i=2;i<=n;i++){if( arr[i]==true ){list.add((long)i);}for(int j=2*i;j<=n;j+=i){arr[j]=false;}}return list;}
   
      public static void printarr(int [] arr){for(int i=0;i<arr.length;i++){pw.print(arr[i]+" ");}}
      public static void printarr(long [] arr){for(int i=0;i<arr.length;i++){pw.print(arr[i]+" ");}}
      
      // int [] arr=sc.intArr(a,b,n);  <b
      static void solve() throws Exception{
          String s=sc.next();
          int c1=0,c=0;
  
          for(int i=0;i<s.length();i++){
              if(s.charAt(i)=='0'){
                  c++;
              }else{
                  c1++;
              }
          }
          if( c1==s.length() ){
              pw.println(0);
              return;
          }else if( c==s.length() ){
              pw.println(1);
              return;
          }
  
          int ans=2,mex=0;
  
          for(int i=0;i<s.length();i++){
              if(s.charAt(i)=='0'){
                  while(i<s.length() && s.charAt(i)=='0'){
                      i++;
                  }
                  i--;
                  mex+=1;
              }
          }
          pw.println(Math.min(mex,ans));
  
  
     }
     
     public static void main(String[] args) throws Exception{
          try {
              System.setIn(new FileInputStream("input.txt"));
              System.setOut(new PrintStream(new FileOutputStream("output.txt")));
          } catch (Exception e) {
              System.err.println("Error");
          }
          sc= new FastReader();
          pw = new PrintWriter(System.out);
  
          int tc=1;
          tc=sc.nextInt();
          for(int ii=1;ii<=tc;ii++) {
  //            pw.printf("Case #%d: "b, i);
              solve();
          }
         
          pw.flush();
   
      
   
      }
   
  }