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 static java.lang.Math.max;
  import static java.lang.Math.min;
  import static java.lang.Math.abs;
  import static java.lang.System.out;
  import java.util.*;
  import java.io.*;
  
  public class A{
      public static void main(String[] args) throws IOException {
          FastScanner s = new FastScanner();
              int t = s.nextInt();
  //        int t = 1;
          while (t-- > 0) {
              String g=s.next();
              if(g.charAt(g.length()-1)=='0')g+=1;
              else g+=0;
              char[] a = g.toCharArray();
              long ans=0;
              for(int i=1;i<a.length;i++){
                  if(a[i]==a[i-1])continue;
                  else{
                      if(a[i-1]=='0')ans++;
                  }
              }
              if(ans==0){
                  if(a[0]=='1') out.println(0);
                  else out.println(1);
                  continue;
              }
              out.println(min(ans,2));
          }
      }
      static long gcd(long a, long b) throws IOException {
          return (b == 0) ? a : gcd(b, a % b);
      }
  
      static int gcd(int a, int b) throws IOException {
          return (b == 0) ? a : gcd(b, a % b);
      }
  
      static void sortr(int[] a) {
          ArrayList<Integer> l = new ArrayList<>();
          for (int i : a) l.add(i);
          Collections.sort(l, Collections.reverseOrder());
          for (int i = 0; i < a.length; i++) a[i] = l.get(i);
      }
  
      static void sort(int[] a) {
          ArrayList<Integer> l = new ArrayList<>();
          for (int i : a) l.add(i);
          Collections.sort(l);
          for (int i = 0; i < a.length; i++) a[i] = l.get(i);
      }
  
      static void sort(long[] a) {
          ArrayList<Long> l = new ArrayList<>();
          for (long i : a) l.add(i);
          Collections.sort(l);
          for (int i = 0; i < a.length; i++) a[i] = l.get(i);
      }
  
      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());
          }
      }
  }
  