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.*;
  
  public class Mainnn {
  
      // static final File ip = new File("input.txt");
      // static final File op = new File("output.txt");
      // static {
      //     try {
      //         System.setOut(new PrintStream(op));
      //         System.setIn(new FileInputStream(ip));
      //     } catch (Exception e) {
      //     }
  
      //     // in = new InputReader(System.in);
      // }
  
      public static void main(String[] args) {
          MyScanner sc = new MyScanner();
          out = new PrintWriter(new BufferedOutputStream(System.out));
  
          // out.println(result); // print via PrintWriter
          /////////////////////////////////////////////
  
          int test = sc.nextInt();
          while (test-- != 0) {
              String s = sc.next();
              s+='*';
              int n = s.length();
              int o=0,z=0;
              for(int i=0;i<n-1;i++)
              {
                 if(s.charAt(i) != s.charAt(i+1) && s.charAt(i)=='0') z++;
                 else if(s.charAt(i) != s.charAt(i+1) && s.charAt(i)=='1') o++;
              }
              System.out.println(Math.min(z, 2));
          }
  
          /////////////////////////////////////////////
          // Stop writing your solution here. -------------------------------------
          out.close();
      }
  
      // Arrays.sort(a, new Comparator<Pair>() {
      // @Override
      // public int compare(Pair p1, Pair p2) {
      // if(p1.st - p2.st > 0) return 1;
      // if(p1.st - p2.st < 0) return -1;
      // return 0;
      // }
      // });
  
      static void sortI(int[] arr) {
          int n = arr.length;
          Random rnd = new Random();
          for (int i = 0; i < n; ++i) {
              int tmp = arr[i];
              int randomPos = i + rnd.nextInt(n - i);
              arr[i] = arr[randomPos];
              arr[randomPos] = tmp;
          }
          Arrays.sort(arr);
      }
  
      static void sortL(long[] arr) {
          int n = arr.length;
          Random rnd = new Random();
          for (int i = 0; i < n; ++i) {
              long tmp = arr[i];
              int randomPos = i + rnd.nextInt(n - i);
              arr[i] = arr[randomPos];
              arr[randomPos] = tmp;
          }
          Arrays.sort(arr);
      }
  
      private static long gcd(long a, long b) {
          return (a == 0 ? b : gcd(b % a, a));
      }
  
      static long power(long x, long y, long p) {
          long res = 1;
          x = x % p;
          if (x == 0)
              return 0;
          while (y > 0) {
              if ((y & 1) != 0)
                  res = (res * x) % p;
              y = y >> 1;
              x = (x * x) % p;
          }
          return res;
      }
  
      // -----------PrintWriter for faster output---------------------------------
      public static PrintWriter out;
  
      // -----------MyScanner class for faster input----------
      public static class MyScanner {
          BufferedReader br;
          StringTokenizer st;
  
          public MyScanner() {
              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;
          }
  
      }
      // --------------------------------------------------------
  }