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.util.*;
  
  public class Solution {
  
      /* -------------------------------------- Main Start ----------------------------------- */
      public static void main(String[] args) throws IOException {
          FastReader fs = new FastReader();
          StringBuilder sb = new StringBuilder();
          int t = Integer.parseInt(fs.nextLine().trim());
          while (t-- > 0) {
              char[] str = fs.nextLine().toCharArray();
              ArrayList<Character> zero = new ArrayList<>();
              char s = str[0];
              int temZ = 0, temO = 0;
              if (str[0] == '0') {
                  temZ++;
              } else {
                  temO++;
              }
              for (int i = 1; i < str.length; i++) {
                  if (str[i] == '0') {
                      temZ++;
                  } else {
                      temO++;
                  }
                  if (s != str[i]) {
                      zero.add(s);
                      s = str[i];
                  }
              }
              if (zero.size() == 0 || s != zero.get(zero.size() - 1)) {
                  zero.add(s);
              }
              int min = 0;
              for (int i = 0; i < zero.size(); i++) {
                  if (zero.get(i) == '0') {
                      min++;
                  }
              }
              int curr = 2;
              if (temZ == 0) {
                  curr = 0;
              } else if (temO == 0) {
                  curr = 1;
              }
              System.out.println(Math.min(min, curr));
          }
      }
  
      /* ---------------------------------------------------------------------------------------------------------------------------------------------------------------------- 
    
        StringBuilder sb = new StringBuilder();
        sb.append(x + "\n");
      
        Collections.sort(arr, (a, b) -> Integer.compare(a[0], b[0]))
        arr.toArray(new int[arr.size()][])
      
       */
      private static final int MOD_1 = 1000000000 + 7;
  
      static class FastReader {
  
          BufferedReader br;
          StringTokenizer st;
  
          public FastReader() throws IOException {
              br = new BufferedReader(new InputStreamReader(System.in));
          }
  
          String next() throws IOException {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int nextInt() throws IOException {
              return Integer.parseInt(next());
          }
  
          long nextLong() throws IOException {
              return Long.parseLong(next());
          }
  
          double nextDouble() throws IOException {
              return Double.parseDouble(next());
          }
  
          String nextLine() throws IOException {
              String str = "";
              try {
                  str = br.readLine();
              } catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
  
          /* --------------------------------------------------------------------------------------------------------------------------------------------------------------------- */
          void printArray(int[] arr) {
              System.out.println(Arrays.toString(arr));
          }
  
          void printArray(String[] arr) {
              System.out.println(Arrays.toString(arr));
          }
  
          void printArray(long[] arr) {
              System.out.println(Arrays.toString(arr));
          }
  
          void print(int data) {
              System.out.println(data);
          }
  
          void print(String data) {
              System.out.println(data);
          }
  
          void print(long data) {
              System.out.println(data);
          }
  
          int[] II(int n) throws IOException {
              int[] d = new int[n];
              String[] arr = nextLine().split(" ");
              for (int i = 0; i < n; i++) {
                  d[i] = Integer.parseInt(arr[i]);
              }
  
              return d;
          }
  
          String[] IS(int n) throws IOException {
              return nextLine().split(" ");
          }
  
          long[] IL(int n) throws IOException {
              long[] d = new long[n];
              String[] arr = nextLine().split(" ");
              for (int i = 0; i < n; i++) {
                  d[i] = Long.parseLong(arr[i]);
              }
              return d;
          }
  
          public long gcd(long a, long b) {
              if (a == 0) {
                  return b;
              }
              return gcd(b % a, a);
          }
  
          public long power(long x, long y, long p) {
              long res = 1;
              x = x % p;
              if (x == 0) {
                  return 0;
              }
  
              while (y > 0) {
                  if ((y & 1) == 1) {
                      res = (res * x) % p;
                  }
                  y = y >> 1;
                  x = (x * x) % p;
              }
              return res;
          }
  
          void sieveOfEratosthenes(boolean prime[], int size) {
              Arrays.fill(prime, true);
              prime[0] = prime[1] = false;
              prime[2] = true;
  
              for (int p = 2; p * p < size; p++) {
                  if (prime[p] == true) {
                      for (int i = p * p; i < size; i += p) {
                          prime[i] = false;
                      }
                  }
              }
          }
  
          public long fact(long n) {
              long ans = 1;
              for (int i = 2; i <= n; i++) {
                  ans = (ans * i) % MOD_1;
              }
              return ans;
          }
  
          public long lcm(long a, long b) {
              return (a * b) / gcd(a, b);
          }
  
          public int log2(int N) {
              int result = (int) (Math.log(N) / Math.log(2));
              return result;
          }
  
          // find first element in the arr that is equal to the x. 
          // If x is not present in arr then it will return first element that is greater than x
          // It return arr.size() if all element are smaller than x
          public int lowerBound(int[] arr, int x) {
              int low = 0, high = arr.length, mid = -1;
  
              while (low < high) {
                  mid = (low + high) / 2;
  
                  if (arr[mid] >= x) {
                      high = mid;
                  } else {
                      low = mid + 1;
                  }
              }
  
              return low;
          }
  
          // find first element in the arr that is greater than x. 
          // It return arr.size() if all element are smaller than x
          public int upperBound(int[] arr, int x) {
              int low = 0, high = arr.length, mid = -1;
  
              while (low < high) {
                  mid = (low + high) / 2;
  
                  if (arr[mid] > x) {
                      high = mid;
                  } else {
                      low = mid + 1;
                  }
              }
  
              return low;
          }
  
      }
  
  }
  