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>
  #pragma GCC optimize("Ofast")
  using namespace std;
  const int N = 100010, INF = 0x3f3f3f3f;
  int T;
  void solve() {
    string s;
    cin >> s;
    int cnt0 = 0, cnt1 = 0, kuai = 0;
    for (auto i : s)
      if (i == '0')
        cnt0++;
      else
        cnt1++;
    for (int i = 0; i < s.size(); i++) {
      if (s[i] == '0' && !i)
        kuai++;
      else if (s[i] == '0' && i && s[i - 1] != '0')
        kuai++;
    }
    int ans = 0;
    if (cnt0 == 0)
      ans = 0;
    else if (cnt1 == 0)
      ans = 1;
    else if (!(kuai - 1))
      ans = 1;
    else
      ans = 2;
    cout << ans << '\n';
  }
  int main(void) {
    clock_t c1 = clock();
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    T = 1;
    cin >> T;
    while (T--) solve();
    return 0;
  }
  