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;
  void __print(int x) { cerr << x; }
  void __print(long long x) { cerr << x; }
  void __print(unsigned long long x) { cerr << x; }
  void __print(long double x) { cerr << x; }
  void __print(char x) { cerr << x; }
  void __print(const char *x) { cerr << x; }
  void __print(const string &x) { cerr << x; }
  void __print(bool x) { cerr << (x ? "true" : "false"); }
  template <typename T, typename V>
  void __print(const pair<T, V> &x) {
    cerr << '{';
    __print(x.first);
    cerr << ", ";
    __print(x.second);
    cerr << '}';
  }
  template <typename T>
  void __print(const T &x) {
    int f = 0;
    cerr << '{';
    for (auto &i : x) {
      cerr << (f++ ? ", " : "");
      __print(i);
    }
    cerr << "}";
  }
  void _print() { cerr << "]\n"; }
  template <typename T, typename... V>
  void _print(T t, V... v) {
    __print(t);
    if (sizeof...(v)) {
      cerr << ", ";
    }
    _print(v...);
  }
  string go(long long i, long long n, long long dist, string nw, long long k,
            string ori) {
    ;
    if (dist > k) {
      return "";
    }
    if (nw.substr(0, i) < ori.substr(0, i)) {
      return "";
    }
    if (i >= n) {
      return nw;
    }
    for (char val = '0'; val <= '9'; ++val) {
      nw[i] = val;
      bool found = false;
      for (long long j = 0; j < i; ++j) {
        if (nw[j] == val) {
          found = true;
          break;
        }
      }
      string chk = go(i + 1, n, dist + (found == false), nw, k, ori);
      if ((long long)chk.size()) {
        return chk;
      }
    }
    return "";
  }
  void solve() {
    long long n, k;
    cin >> n >> k;
    string s = to_string(n);
    long long dist = 0;
    map<char, long long> g;
    for (auto ch : s) {
      g[ch]++;
    }
    string ans =
        go(0, (long long)s.size(), 0, string((long long)s.size(), '0'), k, s);
    cout << ans << '\n';
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    long long t = 1;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  