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>
  #pragma GCC optimize("O3")
  #pragma GCC target("sse4")
  using namespace std;
  template <class T>
  using pq = priority_queue<T>;
  template <class T>
  using pqg = priority_queue<T, vector<T>, greater<T>>;
  template <class T>
  bool ckmin(T& a, const T& b) {
    return b < a ? a = b, 1 : 0;
  }
  template <class T>
  bool ckmax(T& a, const T& b) {
    return a < b ? a = b, 1 : 0;
  }
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  void __print(int x) { cerr << x; }
  void __print(long x) { cerr << x; }
  void __print(long long x) { cerr << x; }
  void __print(unsigned x) { cerr << x; }
  void __print(unsigned long x) { cerr << x; }
  void __print(unsigned long long x) { cerr << x; }
  void __print(float x) { cerr << x; }
  void __print(double 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 first = 0;
    cerr << '{';
    for (auto& i : x) cerr << (first++ ? ", " : ""), __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...);
  }
  const int MOD = 1000000007;
  const char nl = '\n';
  const int MX = 100001;
  int K;
  vector<int> digs;
  int pre;
  set<int> used;
  int go(int rem, bool match) {
    if (rem == 0) {
      return pre;
    }
    int p = (int)(digs).size() - rem;
    int numTried = 0;
    for (int i = 0; i < (10); i++) {
      if (match && i < digs[p]) continue;
      if ((int)(used).size() == K && used.count(i) == 0) {
        continue;
      }
      if (numTried >= 2) continue;
      numTried++;
      pre *= 10;
      pre += i;
      bool curIn = used.count(i);
      if (!curIn) used.insert(i);
      int res = go(rem - 1, match && i == digs[p]);
      if (!curIn) used.erase(i);
      pre /= 10;
      if (res != -1) {
        return res;
      }
    }
    return -1;
  }
  void solve() {
    int N;
    cin >> N >> K;
    digs.clear();
    while (N) {
      digs.push_back(N % 10);
      N /= 10;
    }
    reverse(digs.begin(), digs.end());
    used.clear();
    pre = 0;
    int res = go((int)(digs).size(), true);
    if (res == -1) {
      cout << 1;
      for (int i = 0; i < ((int)(digs).size()); i++) cout << 0;
      cout << nl;
    } else
      cout << res << nl;
  }
  int main() {
    cin.tie(0)->sync_with_stdio(0);
    cin.exceptions(cin.failbit);
    int T = 1;
    cin >> T;
    while (T--) {
      solve();
    }
    return 0;
  }
  