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 int maxn = 200009;
  vector<int> a;
  string in;
  int k;
  string build(string s, vector<char> ch, int ind, bool bigger) {
    char nw = s[ind];
    if (ind == s.size()) return s;
    string tmp;
    if (bigger) {
      for (char c = '0'; c <= '9'; ++c) {
        bool check = true;
        for (int i = 0; i < ch.size() && check; ++i)
          if (ch[i] == c) check = false;
        if (check == false) {
          s[ind] = c;
          tmp = build(s, ch, ind + 1, 1);
          if (tmp != "-1") return tmp;
          s[ind] = nw;
        } else {
          if (ch.size() == k) continue;
          ch.push_back(c);
          s[ind] = c;
          tmp = build(s, ch, ind + 1, 1);
          s[ind] = nw;
          if (tmp != "-1") return tmp;
          ch.pop_back();
        }
      }
    } else {
      for (char c = nw; c <= '9'; ++c) {
        bool check = true;
        for (int i = 0; i < ch.size() && check; ++i)
          if (ch[i] == c) check = false;
        if (check == true) {
          if (ch.size() == k) continue;
          ch.push_back(c);
          s[ind] = c;
          if (c == nw)
            tmp = build(s, ch, ind + 1, 0);
          else
            tmp = build(s, ch, ind + 1, 1);
          if (tmp != "-1") return tmp;
          s[ind] = nw;
          ch.pop_back();
        } else {
          s[ind] = c;
          if (c == nw)
            tmp = build(s, ch, ind + 1, 0);
          else
            tmp = build(s, ch, ind + 1, 1);
          s[ind] = nw;
          if (tmp != "-1") return tmp;
        }
      }
    }
    return "-1";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
      cin >> in >> k;
      vector<char> v;
      cout << build(in, v, 0, 0) << endl;
    }
    return 0;
  }
  