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;
  template <typename T, size_t sz>
  using ar = array<T, sz>;
  template <typename TOStream, typename T1, typename T2>
  inline TOStream &operator<<(TOStream &s, const pair<T1, T2> &p) {
    s << "[" << p.first << ", " << p.second << "]";
    return s;
  }
  template <typename TOStream, typename TContainer>
  inline TOStream &operator<<(TOStream &s, const TContainer &container) {
    s << "[";
    bool first = true;
    for (const auto &val : container) {
      if (!first) {
        s << ", ";
      } else {
        first = false;
      }
      s << val;
    }
    s << "]";
    return s;
  }
  template <typename TIStream, typename T1, typename T2>
  inline TIStream &operator>>(TIStream &s, pair<T1, T2> &p) {
    s >> p.first >> p.second;
    return s;
  }
  template <typename TOStream>
  inline TOStream &operator<<(TOStream &s, const pair<int, int> &p) {
    s << "(" << p.first << ", " << p.second << ")";
    return s;
  }
  bool search(int k, int i, const string &s, const set<char> &digs, string &r,
              set<char> &taken) {
    if (i >= int(s.size())) {
      return true;
    }
    if (k == taken.size()) {
      auto iter = taken.lower_bound(s[i]);
      if (*iter == s[i]) {
        r.push_back(*iter);
        if (search(k, i + 1, s, digs, r, taken)) {
          return true;
        }
        r.pop_back();
        iter++;
      }
      if (iter != taken.end()) {
        r.push_back(*iter);
        for (int j = i + 1; j < int(s.size()); j++) {
          r.push_back(*taken.begin());
        }
        return true;
      }
      return false;
    }
    r.push_back(s[i]);
    taken.insert(s[i]);
    if (search(k, i + 1, s, digs, r, taken)) {
      return true;
    }
    r.pop_back();
    taken.erase(s[i]);
    if (s[i] < '9') {
      r.push_back(s[i] + 1);
      taken.insert(s[i] + 1);
      for (int j = i + 1; j < int(s.size()); j++) {
        r.push_back(taken.size() == k ? *taken.begin() : '0');
      }
      return true;
    }
    return false;
  }
  void solve() {
    string s;
    int k;
    cin >> s >> k;
    string r;
    set<char> digs(s.begin(), s.end());
    set<char> taken;
    search(k, 0, s, digs, r, taken);
    cout << r << endl;
  }
  int main() {
    ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
    int t;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  