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;
  int t;
  int k;
  string s;
  string solve() {
    while (1) {
      set<char> q;
      for (int i = 0; i < s.size(); i++) q.insert(s[i]);
      if (q.size() <= k) return s;
      q.clear();
      for (int pre = 0; pre < s.size(); pre++) {
        q.insert(s[pre]);
        if (q.size() > k) {
          while (s[pre] == '9') pre--;
          s[pre]++;
          for (int i = pre + 1; i < s.size(); i++) {
            s[i] = '0';
          }
          break;
        }
      }
    }
  }
  int main() {
    cin >> t;
    while (t--) {
      cin >> s >> k;
      cout << solve() << endl;
    }
    return 0;
  }
  