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;
  template <class T>
  ostream &operator<<(ostream &os, const vector<T> &p) {
    os << "[ ";
    for (auto &it : p) os << it << " ";
    return os << "]";
  }
  template <class T>
  ostream &operator<<(ostream &os, const unordered_set<T> &p) {
    os << "[ ";
    for (auto &it : p) os << it << " ";
    return os << "]";
  }
  template <class T>
  ostream &operator<<(ostream &os, const set<T> &p) {
    os << "[ ";
    for (auto &it : p) os << it << " ";
    return os << "]";
  }
  template <class T>
  ostream &operator<<(ostream &os, const multiset<T> &p) {
    os << "[ ";
    for (auto &it : p) os << it << " ";
    return os << "]";
  }
  template <class S, class T>
  ostream &operator<<(ostream &os, const pair<S, T> &p) {
    return os << "(" << p.first << ", " << p.second << ")";
  }
  template <class S, class T>
  ostream &operator<<(ostream &os, const unordered_map<S, T> &p) {
    os << "[ ";
    for (auto &it : p) os << it << " ";
    return os << "]";
  }
  template <class S, class T>
  ostream &operator<<(ostream &os, const map<S, T> &p) {
    os << "[ ";
    for (auto &it : p) os << it << " ";
    return os << "]";
  }
  template <class T>
  void dbs(string str, T t) {
    cerr << str << " : " << t << "\n";
  }
  template <class T, class... S>
  void dbs(string str, T t, S... s) {
    long long idx = str.find(',');
    cerr << str.substr(0, idx) << " : " << t << ", ";
    dbs(str.substr(idx + 1), s...);
  }
  long long cel(long long num, long long den) {
    if (num % den == 0)
      return num / den;
    else
      return 1 + num / den;
  }
  long long modExp(long long x, long long y, long long mod) {
    x %= mod, y %= (mod - 1);
    long long res = 1;
    while (y) {
      if (y & 1) res = (res * x) % mod;
      y /= 2, x = (x * x) % mod;
    }
    return res % mod;
  }
  int32_t main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    long long testcases = 1;
    cin >> testcases;
    for (long long testcase = 1; testcase <= testcases; testcase++) {
      string s;
      cin >> s;
      long long n = s.length(), count = 0;
      for (long long i = 0; i < n;) {
        while (i < n and s[i] == '1') i++;
        if (i == n or count >= 2) break;
        count++;
        while (i < n and s[i] == '0') i++;
      }
      cout << min(count, (long long)(2)) << endl;
    }
    return 0;
  }
  