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;
  int main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int t;
    cin >> t;
    while (t--) {
      string n;
      cin >> n;
      int k;
      cin >> k;
      map<char, int> used;
      int l = (int)n.size();
      if (l == 1) {
        cout << n << "\n";
        continue;
      }
      int cnt = 1;
      int pos = 0;
      used[n[0]] = 1;
      char curmx = n[0];
      for (int i = 1; i < l; i++) {
        if (cnt > k) {
          break;
        }
        if (n[i] != '9') {
          if (!used[n[i] + 1] && cnt == k && curmx < n[i] + 1) {
          } else {
            pos = i;
          }
        }
        if (!used[n[i]]) {
          cnt++;
          used[n[i]] = 1;
        }
        curmx = max(curmx, n[i]);
      }
      if (cnt <= k) {
        cout << n << '\n';
        continue;
      }
      set<char> st;
      for (int i = 0; i < pos; i++) {
        cout << n[i];
        st.insert(n[i]);
      }
      if ((int)st.size() == k) {
        cout << *st.lower_bound(n[pos] + 1);
      } else {
        st.insert(n[pos] + 1);
        cout << (char)(n[pos] + 1);
      }
      if ((int)st.size() < k) {
        for (int i = pos + 1; i < l; i++) {
          cout << 0;
        }
      } else {
        for (int i = pos + 1; i < l; i++) {
          cout << *st.begin();
        }
      }
      cout << "\n";
    }
    return 0;
  }
  