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;
  bool run(int i, string &s, string &ans, int k, unordered_map<int, int> &m,
           bool greater) {
    int n = s.length();
    if (i == n) return true;
    int d = greater ? 0 : s[i] - '0';
    while (d < 10) {
      if (m.find(d) != m.end() || m.size() < k) {
        m[d]++;
        ans[i] = d + '0';
        bool f = run(i + 1, s, ans, k, m, greater || (s[i] - '0' < d));
        if (f) return true;
        m[d]--;
        if (m[d] == 0) m.erase(d);
      }
      d++;
    }
    return false;
  }
  int solve(int n, int k) {
    unordered_map<int, int> m;
    int x = log10(n);
    x = pow(10, x);
    string s = to_string(n);
    string ans = to_string(x);
    run(0, s, ans, k, m, false);
    return stoi(ans);
  }
  int main() {
    int t;
    cin >> t;
    while (t--) {
      int n, k;
      cin >> n >> k;
      cout << solve(n, k) << endl;
    }
    return 0;
  }
  