description: It is a complicated version of problem F1. 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 ≤ 10).
  
  Output
  
  For each test case output on a separate line x — the minimum k-beautiful integer such that x ≥ n.
  
  Example
  
  Input
  
  
  6
  2021 3
  177890 2
  34512 3
  724533 4
  998244353 1
  12345678 10
  
  
  Output
  
  
  2021
  181111
  34533
  724542
  999999999
  12345678
solution: #include <bits/stdc++.h>
  using namespace std;
  string s;
  vector<int> ans;
  set<int> st;
  unordered_map<int, int> um;
  bool f(int idx, int k, bool great) {
    if (idx == s.size()) return true;
    int angka = s[idx] - '0';
    if (great) {
      ans.push_back(*st.begin());
      if (f(idx + 1, k, true)) return true;
      return false;
    }
    if (k == 0) {
      auto it = st.begin();
      while (it != st.end()) {
        int val = *it;
        if (val >= angka) {
          ans.push_back(val);
          if (val > angka) {
            if (f(idx + 1, k, true)) return true;
          } else {
            if (f(idx + 1, k, false)) return true;
          }
          ans.pop_back();
        }
        it++;
      }
      return false;
    }
    bool e = st.find(angka) != st.end();
    ans.push_back(angka);
    st.insert(angka);
    um[angka]++;
    if (e) {
      if (f(idx + 1, k, false)) return true;
    } else {
      if (f(idx + 1, k - 1, false)) return true;
    }
    ans.pop_back();
    um[angka]--;
    if (um[angka] == 0) st.erase(angka);
    angka++;
    e = st.find(angka) != st.end();
    ans.push_back(angka);
    st.insert(angka);
    um[angka]++;
    if (e) {
      st.insert(0);
      if (f(idx + 1, k, true)) return true;
    } else {
      if (f(idx + 1, k - 1, true)) return true;
    }
    ans.pop_back();
    um[angka]--;
    if (um[angka] == 0) st.erase(angka);
    return false;
  }
  void solve() {
    int k;
    cin >> s >> k;
    ans.clear();
    st.clear();
    um.clear();
    f(0, k, false);
    for (auto x : ans) cout << x;
    cout << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ;
    int t = 1;
    cin >> t;
    for (int tt = 0; tt < t; tt++) {
      solve();
    }
    return 0;
  }
  