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;
  using vi = vector<int>;
  using vvi = vector<vi>;
  using vvvi = vector<vvi>;
  using ll = long long;
  using vll = vector<ll>;
  using vvll = vector<vll>;
  using vvvll = vector<vvll>;
  using ull = unsigned long long;
  using vb = vector<bool>;
  using vvb = vector<vb>;
  using vd = vector<double>;
  using vvd = vector<vd>;
  using vvvd = vector<vvd>;
  using vs = vector<string>;
  using vvs = vector<vs>;
  using vc = vector<char>;
  using vvc = vector<vc>;
  using ii = pair<int, int>;
  using vii = vector<ii>;
  using vvii = vector<vii>;
  using dd = pair<double, double>;
  using vdd = vector<dd>;
  using vvdd = vector<vdd>;
  using Mii = map<int, int>;
  using vMii = vector<Mii>;
  using Si = set<int>;
  using vSi = vector<Si>;
  using vvSi = vector<vSi>;
  int main() {
    cin.tie(NULL);
    ios_base::sync_with_stdio(false);
    int t;
    cin >> t;
    while (t--) {
      string s;
      cin >> s;
      int n = s.size();
      bool one = true;
      for (char c : s)
        if (c == '0') one = false;
      if (one)
        cout << 0 << '\n';
      else {
        int first = -1;
        for (int i = 0; i < n; ++i) {
          if (s[i] == '0') {
            first = i;
            break;
          }
        }
        int last = -1;
        for (int i = n - 1; i >= 0; --i) {
          if (s[i] == '0') {
            last = i;
            break;
          }
        }
        bool two = false;
        for (int i = first; i <= last; ++i) {
          if (s[i] == '1') two = true;
        }
        cout << (two ? 2 : 1) << '\n';
      }
    }
  }
  