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: /*
  
  Sample Input 1
  
  6
  01
  1111
  01100
  101
  0000
  01010
  
  Sample Output 1
  
  1
  0
  2
  1
  1
  2
  
  */
  import java.io.*;
  import java.util.*;
    
  public class R1566B
  {
      static int MAX(int x, int y){return ((x > y) ? x : y );}
      static int MIN(int x, int y){return ((x < y) ? x : y );}
      static int sum(int[] A){int sum = 0;for(int i=0; i<A.length; i++)sum+=A[i];return (sum);}
      static int gcd(int a, int b){if (b == 0)return a;return gcd(b, a % b);}
      static boolean isPowerOfTwo(int n){if(n==0)return false;return (int)(Math.ceil((Math.log(n) / Math.log(2)))) ==(int)(Math.floor(((Math.log(n) / Math.log(2)))));}
      static int countDigits(long l)
      {
              if (l >= 1000000000000000000L) return 19;
              if (l >= 100000000000000000L) return 18;
              if (l >= 10000000000000000L) return 17;
              if (l >= 1000000000000000L) return 16;
              if (l >= 100000000000000L) return 15;
              if (l >= 10000000000000L) return 14;
              if (l >= 1000000000000L) return 13;
              if (l >= 100000000000L) return 12;
              if (l >= 10000000000L) return 11;
              if (l >= 1000000000L) return 10;
              if (l >= 100000000L) return 9;
              if (l >= 10000000L) return 8;
              if (l >= 1000000L) return 7;
              if (l >= 100000L) return 6;
              if (l >= 10000L) return 5;
              if (l >= 1000L) return 4;
              if (l >= 100L) return 3;
              if (l >= 10L) return 2;
              return 1;
      }
      static void sort2D(int[][] Arr){Arrays.sort(Arr, (a, b) -> Double.compare(a[0], b[0]));}
      public static void main(String[] args) throws IOException
      {
          BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
          if (System.getProperty("ONLINE_JUDGE") == null)
          {
              try 
              {
                  in = new BufferedReader(new FileReader("input.txt"));
                  System.setOut(new PrintStream(new FileOutputStream("output.txt")));
              }
              catch (Exception e) {
                  //System.out.println("Exception");
              }
          }
          int T = Integer.parseInt(in.readLine());
          while( T -- > 0)
          {
              String lines = in.readLine();
              int count = 0;
              int temp = 0;
              int N = lines.length();
              for(int i = 0; i < N; i++)
              {
                  char temp1 = lines.charAt(i);
                  if(temp1 == '0')
                      count = 1;
                  else if(count == 1)
                  {
                      count = 0;
                      temp++;
                  }
              }
              if(lines.charAt(N-1) == '0')
                  temp++;
              System.out.println(MIN(2,temp));
          }
      }
  }