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;
  template <typename T>
  void chkMax(T &first, T second) {
    if (second > first) first = second;
  }
  template <typename T>
  void chkMin(T &first, T second) {
    if (second < first) first = second;
  }
  template <typename T>
  inline void read(T &f) {
    f = 0;
    T fu = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
      if (c == '-') {
        fu = -1;
      }
      c = getchar();
    }
    while (c >= '0' && c <= '9') {
      f = (f << 3) + (f << 1) + (c & 15);
      c = getchar();
    }
    f *= fu;
  }
  const int MAX_N = 2e5 + 10, MAX_M = 1e6 + 10;
  bool ok(string first, int k) {
    sort(first.begin(), first.end());
    int d = unique(first.begin(), first.end()) - first.begin();
    return d <= k;
  }
  void solve() {
    string n;
    int k;
    cin >> n >> k;
    if (ok(n, k)) {
      cout << n << '\n';
      return;
    }
    for (int i = n.size() - 1; i >= 0; i--) {
      for (n[i]++; n[i] <= '9'; n[i]++) {
        for (char j = '0'; j <= '9'; j++) {
          for (int k = i + 1; k < n.size(); k++) {
            n[k] = j;
          }
          if (ok(n, k)) {
            cout << n << '\n';
            goto nex;
          }
        }
      }
    }
  nex:;
    return;
  }
  int main() {
    int T = 1;
    read(T);
    while (T--) {
      solve();
    }
    return 0;
  }
  