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 namespace std;
  const long long int mod = 1e9 + 7;
  const long long int MAX = LLONG_MAX;
  const long long int MIN = LLONG_MIN;
  const long long int mxsize = 2 * 1e6;
  vector<long long int> adj[mxsize];
  long long int vis[mxsize] = {0};
  void dfs(long long int node) {
    vis[node] = 1;
    for (long long int i = 0; i < adj[node].size(); i++) {
      long long int child = adj[node][i];
      if (!vis[child]) {
        dfs(child);
      }
    }
  }
  long long int power(long long int x, long long int y) {
    long long int res = 1;
    while (x) {
      if (x % 2) {
        res = (res * y) % mod;
      }
      x = x / 2;
      y = (y * y) % mod;
    }
    return res % mod;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ;
    int t;
    cin >> t;
    while (t--) {
      string s;
      cin >> s;
      long long int z = 0, o = 0;
      long long int f = 0, k = 0;
      long long int n = s.length();
      for (long long int i = 0; i < n; i++) {
        if (s[i] == '0') {
          z++;
          f = 1;
        } else {
          if (f == 1) {
            k++;
            f = 0;
          }
          o++;
        }
      }
      if (f == 1) k++;
      if (z == 0) {
        cout << 0 << "\n";
      } else if (o == 0) {
        cout << 1 << "\n";
      } else {
        long long int x = min(2 * 1LL, k);
        cout << x << "\n";
      }
    }
    return 0;
  }
  