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;
  vector<int> get_stacked_value(int n) {
    vector<int> result;
    while (n) {
      result.push_back(n % 10);
      n /= 10;
    }
    reverse(result.begin(), result.end());
    return result;
  }
  int get_int(vector<int> s) {
    int res = 0;
    for (int i : s) {
      s.pop_back();
      res *= 10;
      res += i;
    }
    return res;
  }
  bool is_valid(vector<int> &vec, int n, int k) {
    set<int> s;
    for (int i : vec) s.insert(i);
    int val = get_int(vec);
    return (s.size() <= k) && (val >= n);
  }
  int solve(int n, int k) {
    vector<int> n_vec = get_stacked_value(n);
    if (is_valid(n_vec, n, k)) return n;
    for (int i = n_vec.size() - 1; i >= 0; i--) {
      for (int add = 1; add + n_vec[i] < 10; add++) {
        vector<int> tmp_vec;
        tmp_vec = n_vec;
        tmp_vec[i] = (tmp_vec[i] + add);
        for (int j = i + 1; j < n_vec.size(); j++) tmp_vec[j] = 0;
        set<int> s(tmp_vec.begin(), tmp_vec.begin() + i + 1);
        if (s.size() > k)
          continue;
        else if (s.size() < k)
          for (int j = i + 1; j < n_vec.size(); j++) tmp_vec[j] = 0;
        else {
          int val = 9;
          for (int j = 0; j <= i; j++) val = min(val, tmp_vec[j]);
          for (int j = i + 1; j < n_vec.size(); j++) {
            tmp_vec[j] = val;
          }
        }
        const bool valid = is_valid(tmp_vec, n, k);
        if (valid) return get_int(tmp_vec);
      }
    }
  }
  int main() {
    int T;
    cin >> T;
    while (T--) {
      int n, k;
      cin >> n >> k;
      cout << solve(n, k) << endl;
    }
    return 0;
  }
  