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 main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int t;
    cin >> t;
    for (int test = 0; test < t; test++) {
      string s;
      int n;
      cin >> s >> n;
      set<char> st;
      map<char, int> fst;
      int p = 0, lst = -1;
      while (p < int(s.size()) and int(st.size()) <= n) {
        if (st.count(s[p]) == 0) {
          fst[s[p]] = p;
          st.insert(s[p]);
          if (st.size() == n) {
            lst = p;
          }
        }
        p++;
      }
      if (int(st.size()) <= n) {
        cout << s << '\n';
      } else {
        p--;
        st.erase(s[p]);
        if ((*st.rbegin()) > s[p]) {
          s[p] = (*st.upper_bound(s[p]));
          for (p = p + 1; p < s.size(); p++) {
            s[p] = (*st.begin());
          }
        } else {
          int flag = 0;
          while (!flag and p > lst + 1) {
            p--;
            if ((*st.rbegin()) > s[p]) {
              s[p] = (*st.upper_bound(s[p]));
              for (int pp = p + 1; pp < s.size(); pp++) {
                s[pp] = (*st.begin());
              }
              flag = 1;
            }
          }
          if (flag == 0) {
            p = lst;
            st.erase(s[p]);
            if (st.count(s[p] + 1) > 0) {
              s[p]++;
              for (int pp = p + 1; pp < s.size(); pp++) {
                s[pp] = '0';
              }
            } else {
              s[p] += 1;
              st.insert(s[p]);
              for (p = p + 1; p < s.size(); p++) {
                s[p] = (*st.begin());
              }
            }
          }
        }
        cout << s << '\n';
      }
    }
    return 0;
  }
  