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 ll = long long;
  using ld = long double;
  using str = string;
  using pi = pair<int, int>;
  using pl = pair<ll, ll>;
  using ppi = pair<int, pair<int, int>>;
  using ppl = pair<ll, pair<ll, ll>>;
  using vi = vector<int>;
  using vl = vector<ll>;
  using vpi = vector<pair<int, int>>;
  using vvi = vector<vi>;
  using vvpi = vector<vpi>;
  using vpl = vector<pl>;
  using vvl = vector<vl>;
  using vvpl = vector<vpl>;
  using vppl = vector<ppl>;
  using vppi = vector<ppi>;
  using ull = unsigned long long;
  int MOD = 998244353;
  mt19937 rnd(time(0));
  const int N = 2e5 + 10;
  int mul(int a, int b) { return (1ll * a * b) % MOD; }
  int add(int a, int b) {
    int s = (a + b);
    if (s >= MOD) s -= MOD;
    return s;
  }
  int sub(int a, int b) {
    int s = (a + MOD - b);
    if (s >= MOD) s -= MOD;
    return s;
  }
  int po(int a, ll deg) {
    if (deg == 0) return 1;
    if (deg % 2 == 1) return mul(a, po(a, deg - 1));
    int t = po(a, deg / 2);
    return mul(t, t);
  }
  int inv(int n) { return po(n, MOD - 2); }
  void solve() {
    str s;
    cin >> s;
    int ch = 1, ch2 = 0;
    for (int i = 0; i < s.size(); i++) {
      if (s[i] == '0' && ch2 == 0) {
        ch = 0;
        ch2 = 1;
      } else if (s[i] == '0' && ch2 == 1 && s[i - 1] == '1') {
        ch2 = 2;
        break;
      }
    }
    cout << ch2 << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--) solve();
  }
  