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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.StringTokenizer;
  
  public class GR16B {
  
      public static void main(String[] args) {
          FastScanner sc = new FastScanner();
          int t = sc.ni();
          while (t-- > 0) {
              char[] a = sc.next().toCharArray();
              int a0 = 0, a1 = 0;
              char last = '2';
              for (int i = 0; i < a.length; i++) {
                  if (i == 0) {
                      if (a[i] == '0')
                          a0++;
                      else
                          a1++;
                      last = a[i];
                  } else {
                      if (a[i] != last) {
                          if (a[i] == '0') {
                              a0++;
                          } else {
                              a1++;
                          }
                          last = a[i];
                      }
                  }
              }
              if (a0 == 0) {
                  System.out.println(0);
              } else if (a1 == 0) {
                  System.out.println(1);
              } else {
                  System.out.println(Math.min(2, a0));
              }
          }
      }
  
      static class FastScanner {
          BufferedReader br;
          StringTokenizer st;
  
          public FastScanner() {
              br = new BufferedReader(new InputStreamReader(System.in), 32768);
              st = null;
          }
  
          String next() {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int ni() {
              return Integer.parseInt(next());
          }
  
          int[][] graph(int N, int[][] edges) {
              int[][] graph = new int[N][];
              int[] sz = new int[N];
              for (int[] e: edges) {
                  sz[e[0]] += 1;
                  sz[e[1]] += 1;
              }
              for (int i = 0; i < N; i++) {
                  graph[i] = new int[sz[i]];
              }
              int[] cur = new int[N];
              for (int[] e: edges) {
                  graph[e[0]][cur[e[0]]] = e[1];
                  graph[e[1]][cur[e[1]]] = e[0];
                  cur[e[0]] += 1;
                  cur[e[1]] += 1;
              }
              return graph;
          }
  
          int[] intArray(int N, int mod) {
              int[] ret = new int[N];
              for (int i = 0; i < N; i++)
                  ret[i] = ni()+mod;
              return ret;
          }
  
          long nl() {
              return Long.parseLong(next());
          }
  
          long[] longArray(int N, long mod) {
              long[] ret = new long[N];
              for (int i = 0; i < N; i++)
                  ret[i] = nl()+mod;
              return ret;
          }
  
          double nd() {
              return Double.parseDouble(next());
          }
  
          String nextLine() {
              String str = "";
              try {
                  str = br.readLine();
              } catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
      }
  }
  