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;
  int t;
  int count(string a) {
    int vis[10], cnt = 0;
    memset(vis, 0, sizeof(vis));
    for (int i = 0; i < a.size(); i++) {
      int x = a[i] - '0';
      if (!vis[x]) {
        cnt++;
        vis[x] = 1;
      }
    }
    return cnt;
  }
  long long to_int(string s) {
    long long ans = 0;
    for (int i = 0; i < s.size(); i++) {
      ans = ans * 10 + s[i] - '0';
    }
    return ans;
  }
  int main() {
    cin >> t;
    while (t--) {
      string s;
      int k;
      cin >> s >> k;
      long long ans = 0x3f3f3f3f3f3f3f3f;
      if (count(s) <= k) {
        cout << s << '\n';
        continue;
      }
      for (int i = 0; i < s.size(); i++) {
        for (char j = s[i] + 1; j <= '9'; j++) {
          string ss = s;
          ss[i] = j;
          for (int kk = i + 1; kk <= s.size(); kk++) {
            ss[kk] = ss[i];
          }
          if (count(ss) > k) continue;
          for (int kk = i + 1; kk < s.size(); kk++) {
            for (char r = '0'; r <= '9'; r++) {
              string sss = ss;
              sss[kk] = r;
              if (count(sss) <= k) {
                ss[kk] = r;
                break;
              }
            }
          }
          ans = min(ans, to_int(ss));
        }
      }
      cout << ans << '\n';
    }
    return 0;
  }
  