description: It is a simplified version of problem F2. The difference between them is the constraints (F1: k ≤ 2, F2: k ≤ 10).
  
  You are given an integer n. Find the minimum integer x such that x ≥ n and the number x is k-beautiful.
  
  A number is called k-beautiful if its decimal representation having no leading zeroes contains no more than k different digits. E.g. if k = 2, the numbers 3434443, 55550, 777 and 21 are k-beautiful whereas the numbers 120, 445435 and 998244353 are not.
  
  Input
  
  The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
  
  Each test case consists of one line containing two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 2).
  
  Output
  
  For each test case output on a separate line x — the minimum k-beautiful integer such that x ≥ n.
  
  Example
  
  Input
  
  
  4
  1 1
  221 2
  177890 2
  998244353 1
  
  
  Output
  
  
  1
  221
  181111
  999999999
solution: #include <bits/stdc++.h>
  using namespace std;
  using ll = long long;
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;
    cin >> T;
    while (T--) {
      int n, k;
      cin >> n >> k;
      string N = to_string(n);
      string ans = "99999999999999";
      set<char> S;
      for (auto c : N) {
        S.insert(c);
      }
      if ((int)S.size() <= k) {
        cout << N << "\n";
        continue;
      }
      for (int d = 0; d < (int)N.size(); d++) {
        string base = N.substr(0, d);
        for (int v = N[d] - '0' + 1; v <= 9; v++) {
          string s = base + to_string(v);
          set<char> have;
          for (int i = 0; i < d; i++) {
            have.insert(N[i]);
          }
          have.insert((char)(v + '0'));
          if ((int)have.size() > k) {
            continue;
          }
          if ((int)have.size() < k) {
            have.insert('0');
          }
          while (s.size() < N.size()) {
            s += *have.begin();
          }
          assert(s >= N);
          ans = min(ans, s);
        }
      }
      cout << ans << "\n";
    }
    return 0;
  }
  