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>
  const int INF = 1e9 + 5;
  const int MOD = 1e9 + 7;
  const double P = 3.141592;
  const int hsh = 29;
  using namespace std;
  long long binpow(long long a, long long b) {
    long long res = 1;
    while (b > 0) {
      if (b & 1) {
        res = res * a;
      }
      a = a * a;
      b >>= 1;
    }
    return res;
  }
  long long gcd(long long a, long long b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--) {
      string s;
      cin >> s;
      int ans = 0;
      int start = 0, togsgol = s.size() - 1;
      int barg = s.size();
      while (start < s.size() && togsgol >= 0 && barg > 0) {
        bool plus = false;
        for (start; start < s.size();) {
          if (s[start] == '1' && barg > 0) {
            start++;
            barg--;
          } else
            break;
        }
        for (togsgol; togsgol >= 0;) {
          if (s[togsgol] == '1' && barg > 0) {
            togsgol--;
            barg--;
          } else
            break;
        }
        while (barg > 0 && s[start] == '0') {
          start++;
          barg--;
          plus = true;
        }
        if (plus) ans++;
      }
      ans = min(ans, 2);
      cout << ans << '\n';
    }
  }
  