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;
  void write(vector<int> &v) {
    for (auto i : v) cout << i << " ";
    cout << "\n";
  }
  void read(vector<int> &v) {
    for (auto &i : v) cin >> i;
  }
  const int INF = 1e9;
  const int64_t INFF = 1e18;
  const int N = 1e6 + 69;
  int k;
  string s;
  string calc(int index, string ans) {
    set<char> st((ans).begin(), (ans).end());
    if ((int)(st).size() > k) return "";
    if (ans.substr(0, index) < s.substr(0, index)) return "";
    if (index >= s.length()) {
      return ans;
    }
    for (char c = '0'; c <= '9'; c++) {
      ans.push_back(c);
      string pos = calc(index + 1, ans);
      ans.pop_back();
      if (pos.length() == s.length()) return pos;
    }
  }
  void solve() {
    cin >> s;
    cin >> k;
    cout << calc(0, "") << "\n";
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t = 1;
    cin >> t;
    for (int i = 1; i <= t; i++) {
      solve();
    }
    return 0;
  }
  