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.util.*;
  import java.io.*;
  
  public class CRF16B {
  
      // Snippet for Fast Input-Output
  
      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;
          }
      }
  
      static FastReader sc = new FastReader();
      static PrintWriter out = new PrintWriter(System.out);
  
      static int ni() throws IOException {
          return sc.nextInt();
      }
  
      static long nl() throws IOException {
          return sc.nextLong();
      }
  
      static int[][] nai2(int n, int m) throws IOException {
          int a[][] = new int[n][m];
          for (int i = 0; i < n; i++)
              for (int j = 0; j < m; j++)
                  a[i][j] = ni();
          return a;
      }
  
      static int[] nai(int N, int start) throws IOException {
          int[] A = new int[N + start];
          for (int i = start; i != (N + start); i++) {
              A[i] = ni();
          }
          return A;
      }
  
      static Integer[] naI(int N, int start) throws IOException {
          Integer[] A = new Integer[N + start];
          for (int i = start; i != (N + start); i++) {
              A[i] = ni();
          }
          return A;
      }
  
      static long[] nal(int N) throws IOException {
          long[] A = new long[N];
          for (int i = 0; i != N; i++) {
              A[i] = nl();
          }
          return A;
      }
  
      static void close() {
          out.flush();
      }
  
      // ---------------------Main-Code------------------------------------------
  
      public static void main(String[] args) throws Exception {
          int t = ni();
          while (t-- != 0) {
              String str = sc.next();
              int n = str.length();
              if (n == 1 && str.charAt(0) == '1') {
                  System.out.println(0);
                  continue;
              }
              int count01 = 0;
              boolean isOne = false;
              int firstOne = -1;
              int lastOne = -1;
              for (int i = 0; i < n - 1; i++) {
                  char ch1 = str.charAt(i);
                  char ch2 = str.charAt(i + 1);
                  if (ch1 == '1') {
                      if (firstOne == -1)
                          firstOne = i;
                      isOne = true;
                  }
                  if (ch2 == '1') {
                      lastOne = i + 1;
                      isOne = true;
                  }
  
                  if (ch1 != ch2) {
                      count01++;
                  }
              }
  
              if (count01 == 1 && (firstOne == 0 || lastOne == (n - 1))) {
                  System.out.println(1);
                  continue;
              }
  
              if (count01 == 0 && !isOne) {
                  System.out.println(1);
                  continue;
              }
  
              if (count01 == 0 && isOne) {
                  System.out.println(0);
                  continue;
              }
  
              if (count01 == 2 && (firstOne == 0 && lastOne == (n - 1))) {
                  System.out.println(1);
                  continue;
              }
  
              System.out.println(2);
          }
      }
  
  }
  