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;
  string ans, a;
  int n, k;
  bool go(int i, int n, int k, set<int> &s, int gr) {
    if (i == n) return true;
    int val = a[i] - '0';
    if (gr == 1) {
      int x = 0;
      if (s.size() == k) {
        x = (*s.begin());
      }
      ans.push_back((char)(x + '0'));
      int flag = 0;
      if (s.empty() == true || s.find(x) == s.end()) {
        s.insert(x);
      } else {
        flag = 1;
      }
      if (go(i + 1, n, k, s, gr) == true)
        return true;
      else {
        if (flag == 0) {
          s.erase(x);
        }
        ans.pop_back();
        return false;
      }
    }
    if (s.size() == k) {
      if (s.lower_bound(val) == s.end()) {
        return false;
      }
      int ins = (*s.lower_bound(val)) + '0';
      int nextgr = 0;
      nextgr = ((ins - '0') == val) ? 0 : 1;
      ans.push_back(ins);
      if (go(i + 1, n, k, s, nextgr) == true)
        return true;
      else {
        ans.pop_back();
        if (s.upper_bound(val) == s.end())
          return false;
        else {
          ans.push_back((*s.upper_bound(val)) + '0');
          if (go(i + 1, n, k, s, 1) == true)
            return true;
          else {
            ans.pop_back();
            return false;
          }
        }
      }
    }
    int flag = 0;
    ans.push_back(a[i]);
    if (s.empty() == true || s.find(val) == s.end())
      s.insert(val);
    else
      flag = 1;
    if (go(i + 1, n, k, s, gr) == true)
      return true;
    else {
      ans.pop_back();
      if (flag == 0) s.erase(val);
      int flag2 = 0;
      if (s.empty() == true || s.find((val + 1) % 10) == s.end())
        s.insert((val + 1) % 10);
      else
        flag2 = 1;
      ans.push_back((char)((val + 1) % 10 + '0'));
      if (go(i + 1, n, k, s, 1) == true)
        return true;
      else {
        ans.pop_back();
        if (flag2 == 0) s.erase((val + 1) % 10);
        return false;
      }
    }
  }
  void solve() {
    ans = "", a = "";
    cin >> n >> k;
    a = to_string(n);
    set<int> s;
    go(0, a.size(), k, s, 0);
    cout << ans << endl;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout << fixed << setprecision(10);
    int t = 1;
    cin >> t;
    while (t--) solve();
  }
  