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;
  void solve() {
    string N;
    int K;
    cin >> N >> K;
    while (true) {
      set<char> st;
      for (int i = 0; i < N.size(); i++) st.insert(N[i]);
      if (st.size() <= K) {
        cout << N << endl;
        return;
      }
      st.clear();
      int nw = 0;
      for (;; nw++) {
        st.insert(N[nw]);
        if (st.size() > K) {
          while (N[nw] == '9') nw -= 1;
          N[nw]++;
          for (int i = nw + 1; i < N.size(); i++) N[i] = '0';
          break;
        }
      }
    }
  }
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;
    cin >> T;
    while (T--) solve();
  }
  