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>
  bool check(int32_t n, int32_t k) {
    int32_t mask = 0;
    while (n > 0) {
      mask |= 1 << (n % 10);
      n /= 10;
    }
    return __builtin_popcount(mask) <= k;
  }
  int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    int32_t num_tests;
    std::cin >> num_tests;
    for (int32_t t = 0; t < num_tests; t++) {
      int32_t n, k;
      std::cin >> n >> k;
      std::vector<int32_t> digits;
      int32_t n2 = n;
      while (n2 > 0) {
        digits.push_back(n2 % 10);
        n2 /= 10;
      }
      std::reverse(digits.begin(), digits.end());
      int64_t answer = 1;
      while (answer < n) answer = answer * 10 + 1;
      int64_t pref = 0;
      for (int32_t i = 0; i <= digits.size(); i++) {
        for (int32_t p1 = 0; p1 <= 10; p1++)
          for (int32_t p2 = 0; p2 <= 10; p2++) {
            int64_t new_num = pref;
            if (i == 0 && p1 == 0) continue;
            if (new_num >= n && check(new_num, k))
              answer = std::min(answer, new_num);
            new_num = new_num * 10 + p1;
            while (new_num < n) {
              new_num = new_num * 10 + p2;
            }
            if (new_num >= n && check(new_num, k))
              answer = std::min(answer, new_num);
          }
        pref *= 10;
        pref += digits[i];
      }
      std::cout << answer << "\n";
    }
    return 0;
  }
  