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;
  using i64 = long long;
  using u64 = unsigned long long;
  struct debug {
    ~debug() { cerr << endl; }
    template <class c>
    debug& operator<<(c x) {
      cerr << x;
      return *this;
    }
    template <class c, class d>
    debug& operator<<(pair<c, d> x) {
      *this << "(" << x.first << ", " << x.second << ")";
      return *this;
    }
    template <class c>
    debug& operator<<(vector<c> x) {
      *this << "[";
      int f = 0;
      for (auto it : x) {
        *this << (f ? ", " : "");
        *this << it;
        f = 1;
      }
      *this << "]";
      return *this;
    }
  };
  bool f(int pos, string cur, set<int> digits, bool smaller, int k, int len,
         string n) {
    if ((int)digits.size() > k) {
      return false;
    }
    if (pos == len) {
      printf("%s\n", cur.c_str());
      return true;
    }
    for (char dig = '0'; dig <= '9'; dig++) {
      bool flag = smaller;
      if (smaller) {
      } else {
        if (dig < n[pos]) continue;
        if (dig == n[pos]) {
        } else if (dig > n[pos]) {
          flag = true;
        }
      }
      int already = (digits.find(dig - '0') != digits.end());
      digits.insert(dig - '0');
      cur += dig;
      int done = f(pos + 1, cur, digits, flag, k, len, n);
      if (done) {
        return true;
      }
      cur.pop_back();
      if (!already) {
        digits.erase(dig - '0');
      }
    }
    return false;
  }
  void solve() {
    int n, k;
    scanf("%d%d", &n, &k);
    string t = "";
    set<int> s;
    f(0, t, s, 0, k, (int)to_string(n).length(), to_string(n));
  }
  int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
      solve();
    }
  }
  