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>
  const int N = 2e5 + 100;
  using namespace std;
  clock_t startTime;
  double getCurrentTime() {
    return (double)(clock() - startTime) / CLOCKS_PER_SEC;
  }
  template <typename T>
  void print(const T &t) {
    cout << t << '\n';
  }
  template <typename T, typename... F>
  void print(const T &t, const F &...f) {
    cout << t << " ";
    print(f...);
  }
  template <typename T>
  void input_array(vector<T> &arr) {
    for (int i = 0; i < arr.size(); i++) {
      cin >> arr[i];
    }
  }
  template <typename T>
  void print_array(const T &arr, char c = ' ') {
    for (auto x : arr) {
      cout << x << c;
    }
    cout << '\n';
  }
  void file_i_o() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  }
  bool func(string num, int k) {
    set<char> s;
    for (int i = 0; i < num.length(); i++) {
      s.insert(num[i]);
    }
    if (s.size() <= k) {
      return false;
    } else {
      return true;
    }
  }
  int main() {
    file_i_o();
    startTime = clock();
    int t;
    cin >> t;
    while (t--) {
      string n;
      int k;
      cin >> n >> k;
      while (true) {
        set<char> s;
        for (auto c : n) s.insert(c);
        if (s.size() <= k) break;
        s.clear();
        int ptr = 0;
        for (;; ptr++) {
          s.insert(n[ptr]);
          if (s.size() > k) {
            while (n[ptr] == '9') ptr--;
            n[ptr]++;
            for (int i = ptr + 1; i < n.size(); i++) n[i] = '0';
            break;
          }
        }
      }
      cout << n << "\n";
    }
    return 0;
  }
  