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.OutputStream;
  import java.io.IOException;
  import java.io.InputStream;
  import java.io.PrintWriter;
  import java.util.StringTokenizer;
  import java.io.IOException;
  import java.io.BufferedReader;
  import java.io.FileReader;
  import java.io.InputStreamReader;
  import java.io.InputStream;
  
  /**
   * Built using CHelper plug-in
   * Actual solution is at the top
   *
   * @author Khater
   */
  public class Main {
      public static void main(String[] args) {
          InputStream inputStream = System.in;
          OutputStream outputStream = System.out;
          Scanner in = new Scanner(inputStream);
          PrintWriter out = new PrintWriter(outputStream);
          BMINMEXCut solver = new BMINMEXCut();
          solver.solve(1, in, out);
          out.close();
      }
  
      static class BMINMEXCut {
          public void solve(int testNumber, Scanner sc, PrintWriter pw) {
              int t = 1;
              t = sc.nextInt();
              while (t-- > 0) {
                  char[] arr = sc.nextLine().toCharArray();
                  int n = arr.length;
                  int c0 = 0;
                  int c1 = 0;
                  for (char x : arr) {
                      if (x == '0') c0++;
                      else c1++;
                  }
                  int c = 0;
                  for (int i = 0; i < n; i++) {
                      boolean f = false;
                      while (i < n && arr[i] == '0') {
                          if (!f) c++;
                          f = true;
                          i++;
                      }
                  }
                  if (c0 == n) {
                      pw.println(1);
                  } else if (c1 == n) {
                      pw.println(0);
                  } else {
                      pw.println(Math.min(c, 2));
                  }
              }
  
          }
  
      }
  
      static class Scanner {
          StringTokenizer st;
          BufferedReader br;
  
          public Scanner(FileReader r) {
              br = new BufferedReader(r);
          }
  
          public Scanner(InputStream s) {
              br = new BufferedReader(new InputStreamReader(s));
          }
  
          public String next() {
              while (st == null || !st.hasMoreTokens()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      throw new RuntimeException(e);
                  }
              }
              return st.nextToken();
          }
  
          public int nextInt() {
              return Integer.parseInt(next());
          }
  
          public String nextLine() {
              try {
                  return br.readLine();
              } catch (Exception e) {
                  throw new RuntimeException(e);
              }
          }
  
      }
  }
  
  