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>
  int64_t n, k, i;
  std::string s, y;
  std::set<char> st;
  std::string solve() {
    st.clear();
    n = s.length();
    y = "";
    for (i = 0; i < n && st.size() <= k; ++i) {
      y += s[i];
      st.insert(s[i]);
    }
    if (i == n && st.size() <= k) return y;
    s = std::to_string(std::stoi(y) + 1) + std::string(n - y.size(), '0');
    return solve();
  }
  int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::cout.tie(nullptr);
    int t;
    std::cin >> t;
    while (t--) {
      std::cin >> s >> k;
      std::cout << solve() << '\n';
    }
    return 0;
  }
  