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.lang.*;import java.util.*;
  //* --> number of prime numbers less then or equal to x are  -->  x/ln(x)
  //* --> String concatenation using the + operator within a loop should be avoided. Since the String object is immutable, each call for concatenation will
  // result in a new String object being created.
  // THE SIEVE USED HERE WILL RETURN A LIST CONTAINING ALL THE PRIME NUMBERS TILL N
   public class min_mx {static FastScanner sc;static PrintWriter pw;static class FastScanner {InputStreamReader is;BufferedReader br;StringTokenizer st;
  public FastScanner() {is = new InputStreamReader(System.in);br = new BufferedReader(is);}
  String next() throws Exception {while (st == null || !st.hasMoreElements())st = new StringTokenizer(br.readLine());
  return st.nextToken();}int nextInt() throws Exception {return Integer.parseInt(next());}long nextLong() throws Exception {
  return Long.parseLong(next());}int[] readArray(int num) throws Exception {int arr[]=new int[num];
  for(int i=0;i<num;i++)arr[i]=nextInt();return arr;}String nextLine() throws Exception {return br.readLine();
  }} public static boolean power_of_two(int a){if((a&(a-1))==0){ return true;}return false;}
  static boolean PS(double x){if (x >= 0) {double i= Math.sqrt(x);if(i%1!=0){
  return false;}return ((i * i) == x);}return false;}public static int[] ia(int n){int ar[]=new int[n];
  return ar;}public static long[] la(int n){long ar[]=new long[n];return ar;}
  public static void print(int ans,int t){System.out.println("Case"+" "+"#"+t+":"+" "+ans);}
  static long mod=1000000007;static int max=Integer.MIN_VALUE;static int min=Integer.MAX_VALUE;
  public static void sort(long[] arr){//because Arrays.sort() uses quicksort which is dumb
  //Collections.sort() uses merge sort
  ArrayList<Long> ls = new ArrayList<Long>();for(long x: arr)ls.add(x);Collections.sort(ls);
  for(int i=0; i < arr.length; i++)arr[i] = ls.get(i);}public static long fciel(long a, long b) {if (a == 0)  return 0;return (a - 1) / b + 1;}
   static boolean[] is_prime = new boolean[1000001];static ArrayList<Integer> list = new ArrayList<>();
  static long n = 1000000;public static void sieve() {Arrays.fill(is_prime, true);
  is_prime[0] = is_prime[1] = false;for (int i = 2; i * i <= n; i++) {
  if (is_prime[i]) {for (int j = i * i; j <= n; j += i)is_prime[j] = false;}}for (int i = 2; i <= n; i++) {
  if (is_prime[i]) {list.add(i);}}}
  public static void main(String args[]) throws java.lang.Exception {
  sc = new FastScanner();pw = new PrintWriter(System.out);StringBuilder s = new StringBuilder();
  int t=sc.nextInt();
  while(t-->0)
  {
      String str=sc.next();
      int count0=0;
    int i=0;
    while(i<str.length())
    {
          if(str.charAt(i)=='0')
          {
              int j=i;
              while(j<str.length()&&str.charAt(j)=='0')
              {
                  j++;
              }
              count0++;
              i=j;
          }
          
          i++;
      }
   if(count0>2)
   {
       s.append(2);
   }
   else if(count0<=2)
   {
       s.append(count0);
   }
    
      if(t>0)
      {
          s.append("\n");
      }
          }
  pw.print(s);pw.close();}}