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;
  bool ok(string s, int d) {
    sort(s.begin(), s.end());
    return unique(s.begin(), s.end()) - s.begin() <= d;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
      string s;
      int d;
      cin >> s >> d;
      if (ok(s, d)) {
        cout << s << endl;
        continue;
      }
      for (int i = (int)s.size() - 1; i >= 0; i--) {
        for (s[i]++; s[i] <= '9'; s[i]++) {
          for (char j = '0'; j <= '9'; j++) {
            for (int k = i + 1; k < (int)s.size(); k++) s[k] = j;
            if (ok(s, d)) {
              cout << s << endl;
              goto next;
            }
          }
        }
        s[i] = '0';
      }
    next:;
    }
  }
  