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;
  char s[12];
  int len, ans, k;
  int cal(int mask) { return __builtin_popcount(mask); }
  void dfs(int lim, int pos, int mask, int sum) {
    if (pos == len + 1) {
      ans = sum;
      return;
    }
    int start = 0;
    if (lim) {
      start = s[pos];
    }
    for (int i = start; i <= 9; i++) {
      if (ans == 0 && cal(mask | (1 << i)) <= k) {
        dfs(lim && i == start, pos + 1, mask | (1 << i), sum * 10 + i);
      }
    }
  }
  int main() {
    int t;
    int n;
    cin >> t;
    while (t--) {
      scanf("%s", s + 1);
      cin >> k;
      len = strlen(s + 1);
      for (int i = 1; i <= len; i++) {
        s[i] -= '0';
      }
      ans = 0;
      dfs(1, 1, 0, 0);
      cout << ans << endl;
    }
    return 0;
  }
  