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;
  const int N = (int)5e+3 + 8;
  string cal(int i, string n, int dist, int k, string nw) {
    if (dist > k) {
      return "";
    }
    if ((nw.substr(0, i)) < (n.substr(0, i))) {
      return "";
    }
    if (i >= ((int)(n).size())) {
      return nw;
    }
    for (char c = '0'; c <= '9'; c++) {
      bool found = false;
      nw[i] = c;
      for (int j = 0; j < i; j++) {
        if (nw[j] == c) {
          found = true;
          break;
        }
      }
      string ch = cal(i + 1, n, dist + (found == false), k, nw);
      if (((int)(ch).size())) return ch;
    }
    return "";
  }
  void MAIN(long long tc) {
    string n;
    long long k;
    cin >> n >> k;
    string ans = cal(0, n, 0, k, string(((int)(n).size()), '0'));
    cout << ans << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    cout << fixed;
    cout << setprecision(10);
    int test__cases = 1;
    cin >> test__cases;
    for (int i = 1; i <= test__cases; i++) {
      MAIN(i);
    }
  }
  