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;
  signed main() {
    ios::sync_with_stdio(false);
    int t;
    cin >> t;
    while (t--) {
      string str;
      int k;
      cin >> str >> k;
      while (true) {
        set<char> st;
        for (int i = 0; i < str.size(); i++) st.insert(str[i]);
        if (st.size() <= k) break;
        st.clear();
        for (int i = 0; i < str.size(); i++) {
          st.insert(str[i]);
          if (st.size() > k) {
            int pos = i;
            while (str[pos] == '9') pos--;
            str[pos] += 1;
            for (int j = pos + 1; j < str.size(); j++) {
              str[j] = '0';
            }
            break;
          }
        }
      }
      cout << str << endl;
    }
    return 0;
  }
  