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: #include <bits/stdc++.h>
  using namespace std;
  #pragma GCC target("sse4.2")
  void err(istringstream &ss) { cout << "\n"; }
  template <typename T, typename... Args>
  void err(istringstream &ss, const T &a, const Args &...args) {
    string token;
    getline(ss, token, ',');
    cout << token << " = " << a << "; ";
    err(ss, args...);
  }
  int log(int x) { return 32 - __builtin_clz(x) - 1; }
  void read() {}
  template <typename T, typename... Args>
  void read(T &t, Args &...args) {
    cin >> t;
    read(args...);
  }
  struct Complex {
    double x, y;
    Complex(double x = 0, double y = 0) : x(x), y(y) {}
    Complex operator+(const Complex &a) { return Complex(x + a.x, y + a.y); }
    Complex operator-(const Complex &a) { return Complex(x - a.x, y - a.y); }
    Complex operator*(const Complex &a) {
      return Complex(x * a.x - y * a.y, x * a.y + y * a.x);
    }
  };
  bool check_palindrome(string s, int i, int n) {
    if (i >= n) return true;
    return (s[i] == s[n] && check_palindrome(s, i + 1, n - 1));
  }
  long long binpow(long long a, long long b, long long m) {
    a %= m;
    long long res = 1;
    while (b > 0) {
      if (b & 1) res = res * a % m;
      a = a * a % m;
      b >>= 1;
    }
    return res;
  }
  int d[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
  void solve() {
    string s;
    cin >> s;
    if (count(s.begin(), s.end(), '0') == 0) {
      cout << 0 << "\n";
    } else {
      if (int(s.rfind('0') - s.find('0') + 1) == count(s.begin(), s.end(), '0'))
        cout << 1 << "\n";
      else
        cout << 2 << "\n";
    }
  }
  int main() {
    ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
    int t = 1;
    cin >> t;
    for (int c = 1; c <= t; c++) {
      solve();
    }
    return 0;
  }
  