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;
  const long long N = 200010;
  long long n, a[N];
  signed main(void) {
    ios::sync_with_stdio(false);
    cin.tie(0);
    long long T;
    cin >> T;
    while (T--) {
      string n;
      long long k;
      cin >> n >> k;
      while (true) {
        set<char> s;
        for (long long i = 0; i < n.size(); i++) s.insert(n[i]);
        if (s.size() <= k) {
          cout << n << endl;
          break;
        } else {
          s.clear();
          long long pos = 0;
          while (true) {
            s.insert(n[pos]);
            if (s.size() > k) {
              while (n[pos] == '9') pos--;
              n[pos]++;
              for (long long i = pos + 1; i < n.size(); i++) n[i] = '0';
              break;
            }
            pos++;
          }
        }
      }
    }
    return 0;
  }
  