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 mod = 1e9 + 7;
  const long long INF = LLONG_MAX;
  const long long NINF = LLONG_MIN;
  void solve() {
    long long num, k;
    cin >> num >> k;
    vector<long long> d;
    while (num) {
      d.push_back(num % 10);
      num /= 10;
    }
    reverse(d.begin(), d.end());
    long long n = d.size();
    while (1) {
      set<long long> s;
      long long in;
      for (long long i = 0; i < n; i++) {
        s.insert(d[i]);
        if (s.size() > k) {
          in = i;
          break;
        }
      }
      if (s.size() <= k) break;
      while (d[in] == 9) in--;
      d[in]++;
      for (long long i = in + 1; i < n; i++) d[i] = 0;
    }
    for (long long i = 0; i < n; i++) cout << d[i];
    cout << endl;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long t = 1;
    cin >> t;
    while (t-- > 0) {
      solve();
    }
  }
  